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Conic sections

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(Redirected from Hyperbola)

Conic sections are curves created by the intersection of a plane and a cone. There are six types of conic section: the circle, ellipse, hyperbola, parabola, a pair of intersecting straight lines and a single point.

All conics (as they are known) have at least two foci, although the two may coincide or one may be at infinity. They may also be defined as the locus of a point moving between a point and a line, a directrix, such that the ratio between the distances is constant. This ratio is known as "e", or eccentricity.

Ellipses

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Animation showing distance from the foci of an ellipse to several different points on the ellipse.

An ellipse is a locus where the sum of the distances to two foci is kept constant. This sum is also equivalent to the major axis of the ellipse - the major axis being longer of the two lines of symmetry of the ellipse, running through both foci. The eccentricity of an ellipse is less than one.

In Cartesian coordinates, if an ellipse is centered at (h,k), the equation for the ellipse is

   (equation 1)

The lengths of the major and minor axes (also known as the conjugate and transverse) are "a" and "b" respectively.

Exercise 1. Derive equation 1.    (hint)

A circle circumscribed about the ellipse, touching at the two end points of the major axis, is known as the auxiliary circle. The latus rectum of an ellipse passes through the foci and is perpendicular to the major axis.

From a point P(, ) tangents will have the equation:

And normals:

Likewise for the parametric coordinates of P, (a , b ),

Properties of Ellipses

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S and S' are typically regarded as the two foci of the ellipse. Where , these become (ae, 0) and (-ae, 0) respectively. Where these become (0, be) and (0, -be) respectively.

A point P on the ellipse will move about these two foci ut

Where a > b, which is to say the Ellipse will have a major-axes parallel to the x-axis:

The directrix will be:

Circles

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A circle is a special type of the ellipse where the foci are the same point.

Hence, the equation becomes:

Where 'r' represents the radius. And the circle is centered at the origin (0,0)

Hyperbolas

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A special case where the eccentricity of the conic shape is greater than one.

Centered at the origin, Hyperbolas have the general equation:

A point P on will move about the two foci ut

The equations for the tangent and normal to the hyperbola closely resemble that of the ellipse.

From a point P(, ) tangents will have the equation:

And normals:

The directrixes (singular directrix) and foci of hyperbolas are the same as those of ellipses, namely directrixes of and foci of

The asymptotes of a hyperbola lie at

Figure 1: Hyperbola at origin with transverse axis horizontal.

Origin at point .
Foci are points
Vertices are points
Line segment is the

In cartesian geometry in two dimensions hyperbola is locus of a point that moves relative to two fixed points called The distance from one to the other is non-zero. The absolute difference of the distances from point to foci is constant.


See figure 1.


Center of hyperbola is located at the origin and the foci are on the at distance from


has coordinates has coordinates . Line segments


Each point where the curve intersects the transverse axis is called a are the vertices of the ellipse.


By definition


the length of the

Equation of hyperbola at origin

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Transverse axis horizontal

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Figure 1: Hyperbola at origin with transverse axis horizontal.


Origin at point .
Foci are points
Vertices are points

Let point have coordinates



Make appropriate substitutions, expand and result is:

Simplify, gather like terms and recall that, for hyperbola,

Let


Curve in diagram has:

equation or

where

Transverse axis vertical

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Figure 1: Hyperbola at origin with transverse axis vertical.


Origin at point .
Foci are points
Vertices are points

Let point have coordinates

Let Focus 1, have coordinates

Let Focus 2, have coordinates



Make substitutions as above, expand and result is:

Simplify, gather like terms and recall that, for hyperbola,

Let


Curve in diagram has:

equation or

where

Transverse axis oblique

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Figure 1: Hyperbola at origin with transverse axis oblique.


Center of hyperbola is at origin,
Foci are points

With transverse axis oblique the two foci are defined as:

where both are non-zero.

Distance from center to focus and

Let

Let

By definition:

If you make the substitutions as before, result is:

where:


Curve in diagram has:

  • Foci at (24,7), (-24,-7)
  • a = 20
  • equation

Implementation

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# python code

import decimal
dD = decimal.Decimal                        # Decimal object is like float with (almost) infinite precision.
dgt = decimal.getcontext()
Precision = dgt.prec = 28                   # Adjust as necessary.
Tolerance = dD("1e-" + str(Precision-2))    # Adjust as necessary.
# 
# sum_zero(input) is function that calculates sum of all values in input.
# However, if sum is non-zero but very close to 0 and Tolerance permits,
# sum is returned as 0.
# For example sum of (2, -1.99999_99999_99999_99999_99999_99) is
# returned as 0.
#

def hyperbola_ABCF(a,pq, flag = 0) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
for hyperbola at origin, D = E = 0.
ABCF = hyperbola_ABCF( a, pq [, flag] )
(2)a is length of transverse axis.
(p,q) are one focus. Other focus is (-p,-q)
    '''
    thisName = 'hyperbola_ABCF(a,pq, {}) :'.format(bool(flag))
    p,q = pq
    a,p,q = [ dD(str(v)) for v in (a,p,q) ]
    if a == 0 :
        print (thisName)
        print ('  For hyperbola, a must be non-zero.')
        return None
    aa,pp,qq = a**2, p**2, q**2
    cc = pp + qq
    # c = cc.sqrt() and (2)c is distance between foci.
    if cc > aa : pass
    else :
        print (thisName)
        print ('  For hyperbola, c must be > a.')
        return None
    A = aa - pp
    B = aa - qq
    C = -2*p*q
    # F = aa*( pp + qq - aa )
    F = aa*( cc - aa ) # F = aa*bb
    if flag :
        # Print results:
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({}) = 0'
        str2 = str1.format(A,B,C,F)
        print (str2)

        # Equation of small circle, used to display points on grapher.
        str3 = '({},{})    (x - ({}))^2 + (y - ({}))^2 = 1'
        print ('    F1:', str3.format(p,q, p,q))
        print ('    F2:', str3.format(-p,-q, -p,-q))
        print ('    axis: ({})y = ({})x'.format(p,q))

        print ('    aa,a =', aa,a)
        bb = cc-aa ; b = bb.sqrt()
        print ('    bb,b =', bb,b)
        c = cc.sqrt()
        print ('    cc,c =', cc,c)
        if C == 0 :
            # Display intercept form of equation.
            if F > 0 : A,B,C,F = [ -v for v in (A,B,C,F) ]
            str1a = '({})x^2 + ({})y^2 + ({}) = 0'
            str4 = str1a.format(A,B,F)
            print ('   ', str4)

            if (A == bb) and (B == -aa) :
                # (225)x^2 + (-400)y^2 + (-90000) = 0
               str5 = 'xx/(({})^2) - yy/(({})^2) = 1'
               top1_ = 'x^2' ; top2_ = 'y^2'
            elif (A == -aa) and (B == bb) :
                # (-400)x^2 + (225)y^2 + (-90000) = 0
               str5 = 'yy/(({})^2) - xx/(({})^2) = 1'
               top1_ = 'y^2' ; top2_ = 'x^2'
            else : ({}[2])
            str6 = str5.format(a, b)
            print ('   ',str6)

            str5 = '\\frac{{ {} }}{{ {}^2 }} - \\frac{{ {} }}{{ {}^2 }} = 1'
            print('   ','<math>', str5.format(top1_,a, top2_,b), '</math>')
            
            bottom1,bottom2 = [ '({})^2'.format(v) for v in (a,b) ]
            len1,len2 = [ len(v) for v in (bottom1,bottom2) ]
            len1a = (len1-3) >> 1 ; len1b = (len1-3)-len1a
            len2a = (len2-3) >> 1 ; len2b = (len2-3)-len2a
            top1 = '{}{}{}'.format( ' '*len1a,top1_, ' '*len1b )
            top2 = '{}{}'.format( ' '*len2a,top2_ )
            print ( '   ', top1, ' ', top2 )
            print ( '   ', '-'*len1,'-', '-'*len2, '= 1' )
            print ( '   ', bottom1, ' ', bottom2 )
    return A,B,C,F

Examples

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Reversing the process

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The expression "reversing the process" means calculating the values of when given equation of hyperbola at origin, the familiar values

Consider the equation of a hyperbola at origin: This is a hyperbola where

This hyperbola may be expressed as or or where is any real, non-zero number.

However, when this hyperbola is expressed as this format is the hyperbola expressed in "standard form," a notation that greatly simplifies the calculation of

Modify the equations for slightly:

There are four simultaneous equations with four known values and four unknown:

For

For the ideal condition.

Implementation
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# python code

def hyperbola_a_pq (ABCF, flag = 0) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
Here D = E = 0.
a,(p,q) = hyperbola_a_pq (ABCF [, flag])
ABCF implies hyperbola with center at origin (0,0).
if flag : print info about hyperbola.
    '''
    thisName = 'hyperbola_a_pq (ABCF, {}) :'.format(bool(flag))
    A1,B1,C1,F1 = [ dD(str(v)) for v in ABCF ]
    K = 4*F1/(C1**2 - 4*A1*B1)
    A,B,C,F = [ K*v for v in (A1,B1,C1,F1) ] # Standard form.
    # 
    # A = aa - pp ; pp = aa-A
    # B = aa - qq ; qq = aa - B
    # F = aa(pp + qq - aa)
    # F = aa((aa-A) + (aa-B) - aa)
    # F = aaaa-aaA + aaaa-aaB - aaaa
    # F = aaaa - aaA - aaB
    # aaaa - aaA - aaB - F = 0
    # aaaa -(A + B)aa - F = 0
    # 
    # We have a quadratic function in aa.
    # (a_)(aa)(aa) + (b_)(aa) + (c_) = 0
    # Coefficients of quadratic function:
    a_,b_,c_ = 1, -(A+B), -F
    discr = b_**2 - 4*a_*c_
    root = discr.sqrt()
    aa,X = (-b_ + root)/2, (-b_ - root)/2
    # X positive: ellipse
    # X negative: hyperbola
    if X < 0 : pass
    else :
        print (thisName)
        print ('    For hyperbola, X must be < 0. ',)
        return None
    
    a = aa.sqrt() ; pp = aa - A ; p = pp.sqrt()
    if p : q = -C/(2*p)
    else :
        qq = aa - B ; q = qq.sqrt()

    if flag :
        # Print results.
        print ()
        print (thisName)
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({}) = 0'
        str2 = str1.format(A1,B1,C1,F1)
        print (str2)
        if K != 1 :
            str2a = str1.format(A,B,C,F)
            print (str2a, 'Standard form.')
            
        str3 = '({}, {})    (x - ({}))^2 + (y - ({}))^2 = 1'
        print ('    F1:', str3.format(p,q, p,q))
        print ('    F2:', str3.format(-p,-q, -p,-q))
        cc = pp + q**2 ; c = cc.sqrt()
        bb = cc - aa ; b = bb.sqrt()
        for x in 'K A B C F X a b c'.split() :
            print ('   ', x, '=', eval(x))

    return a,(p,q)

Second definition of hyperbola

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Graph of hyperbola where .
At vertex
At point

Hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant.


Let where:

  • is non-zero,

Therefore,

Let directrix have equation where


At vertex


At point


At point


Aim of this section is to prove that :

Statement has been proved for two specific points, vertex and point


Section under "Proof" below proves that statement (3) is true for any point on hyperbola.

Proof

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Proving that .
Graph is part of curve
At point
distance to Directrix2
base =

As expressed above in statement second definition of hyperbola states that hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant.


This section proves that this definition is true for any point on hyperbola.


At point

base


Similar calculations can be used to prove the case for Focus1 and Directrix1 in which case:

Therefore: where


Second definition of hyperbola has been proven valid:

Hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant, called eccentricity

The general hyperbola

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In the two dimensional space of Cartesian Coordinate Geometry the hyperbola may be located anywhere and with any orientation.

To keep the calculation of the general hyperbola as simple as possible, there are two functions that will become very useful:

# python code

def move_section_relative (ABCDEF, gh, flag = 0) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
A,B,C,D1,E1,F1 = move_section_relative ( (A,B,C,D,E,F), (g,h) [, flag]) 
This function moves conic section from its present position to a new
position (g,h) relative to present position.
    '''
    A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]
    g,h = [ dD(str(v)) for v in gh ]
    # 
    # After move, equation of hyperbola becomes:
    # A(x-g)^2 + B(y-h)^2 + C(x-g)(y-h) + D(x-g) + E(y-h) + F = 0
    # or
    # Ax^2 + By^2 + Cxy + (D1)x + (E1)y + (F1) = 0 where:
    D1 = D - C*h - 2*A*g
    E1 = E - C*g - 2*B*h
    F1 = A*g*g + C*g*h + B*h*h + F - D*g - E*h 
    if flag :
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'
        str2 = str1.format(A,B,C,D,E,F)
        print ('from:', str2)
        str3 = str1.format(A,B,C,D1,E1,F1)
        print ('to  :',str3)
    return A,B,C,D1,E1,F1
# python code

def center_of_hyperbola (ABCDEF) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
G,H = center_of_hyperbola (ABCDEF)
    '''
    A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]
    # 
    # If center of hyperbola (G,H) is known then:
    # move_section_relative (ABCDEF, (-G,-H)) 
    # will move hyperbola to origin and D1 = E1 = 0.
    # Equations for D1,E1 become:
    # 
    # D - C*(-H) - 2*A*(-G) = 0
    # E - C*(-G) - 2*B*(-H) = 0
    # 
    # Two simultaneous equations in G,H:
    # 2AG +  CH + D = 0
    #  CG + 2BH + E = 0
    # 
    # C2AG +   CCH +  CD = 0
    # 2ACG + 2A2BH + 2AE = 0
    #        (CC-4AB)H + (CD - 2AE) = 0
    #        (CC-4AB)H = (2AE - CD)
    # 
    #     (2AE - CD)
    # H = ----------
    #     (CC - 4AB)
    #
    H = (2*A*E - C*D)/(C*C - 4*A*B)
    if A :
        G = -(C*H + D)/(2*A)
        return G,H
    if C :
        G = -(2*B*H + E)/C
        return G,H
    ({}[11])

Compare the two simultaneous equations for with those derived from calculus under Slope of curve.

Deriving equation

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To calculate the general equation three values must be known:

  • Focus1
  • Focus2
  • Length of transverse axis or two_a.

Calculate center of hyperbola

Calculate

Calculate equation of hyperbola at origin

Move hyperbola from origin to point

# python code

def hyperbola_ABCDEF (two_a, F1, F2, flag = 0) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0

A,B,C,D,E,F = hyperbola_ABCDEF (two_a, F1, F2  [, flag])
two_a is length of transverse axis
F1 is Focus1: (F1x,F1y)
F2 is Focus2: (F2x,F2y)
if flag : print info about hyperbola.
    '''
    thisName = 'hyperbola_ABCDEF (two_a, F1, F2, {}) :'.format(bool(flag))

    F1x,F1y = F1 ; F2x,F2y = F2
    two_a,F1x,F1y,F2x,F2y = [ dD(str(v)) for v in (two_a,F1x,F1y,F2x,F2y) ]
    
    a = two_a/2
    if a == 0 :
        print (thisName)
        print ('    For hyperbola a must be non-zero.')
        return None

    G = (F1x + F2x)/2 ; H = (F1y + F2y)/2
    p = F2x - G ; q = F2y - H
    aa,pp,qq = a**2, p**2, q**2
    cc = pp + qq
    if cc > aa : pass
    else :
        print (thisName)
        print ('    For hyperbola c must be greater than a.')
        return None
    A00,B00,C00,F00 = hyperbola_ABCF(a, (p,q)) # Hyperbola at origin.
    A,B,C,D,E,F = move_section_relative ( (A00,B00,C00,0,0,F00), (G,H) )
    ( {A==A00, B==B00, C==C00} == {True} ) or ({}[2])
    
    if flag :
        print()
        print(thisName)
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'
        print (str1.format(A,B,C,D,E,F))
        print ('   ', str1.format(A,B,C,0,0,F00))

        str3 = '({},{})    (x - ({}))^2 + (y - ({}))^2 = 1'
        print ('    F1:', str3.format(F1x,F1y, F1x,F1y))
        print ('    F2:', str3.format(F2x,F2y, F2x,F2y))
        print ('    GH:', str3.format(G,H, G,H))
        print ('    f1:', str3.format(p,q, p,q))
        print ('    f2:', str3.format(-p,-q, -p,-q))

        bb = cc - aa
        b = bb.sqrt() ; c = cc.sqrt()
        print ('    a =', a)
        print ('    b =', b)
        print ('    c =', c)
        print ('    eccentricity e =', c/a)
        # Axis:
        Dx = F1x-F2x ; Dy = F1y - F2y
        # Dy x - Dx y + c = 0
        a,b = Dy, -Dx
        # ax + by + c = 0
        c = -(a*F1x + b*F1y)
        print ( '    axis: ({})x + ({})y + ({}) = 0'.format(a,b,c) )
    return A,B,C,D,E,F

Examples

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Curve in Figure 1 below is defined by:

  • Focus1,
  • Focus2,
  • Constant,

Curve has equation:


Curve in Figure 2 below is defined by:

  • Focus1,
  • Focus2,
  • Constant,

Curve has equation:


Curve in Figure 3 below is defined by:

  • Focus1,
  • Focus2,
  • Constant,

Curve has equation:

Reversing the process

[edit | edit source]

The expression "reversing the process" means calculating the values of and length of transverse axis when given equation of general hyperbola, the familiar values


Calculate center of hyperbola.

Move hyperbola to origin.

Calculate a,(p,q) of hyperbola at origin.

Calculate and length of transverse axis,

Implementation
[edit | edit source]
# python code

def hyperbola_2a_F1_F2 (ABCDEF, flag = 0) :
    '''
Ax^2 + By^2 + Cxy + Dx + Ey + F = 0

two_a, F1, F2 = hyperbola_2a_F1_F2 (ABCDEF  [, flag])
two_a, (2)a, is length of transverse axis.
F1, (F1x,F1y), is Focus1.
F2, (F2x,F2y), is Focus2.
#bb = b**2 where (2)b is length of conjugate axis.
if flag == 1 : Check calculations.
if flag == 2 : Check and print result of calculations.
    '''
    thisName = 'hyperbola_2a_F1_F2 (ABCDEF, flag = {}) :'.format(flag)
    ( 2 >= flag >= 0 ) or ({}[1])
    A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]
    G,H = center_of_hyperbola ((A,B,C,D,E,F)) 
    # Move hyperbola to origin.
    A0,B0,C0,D0,E0,F0 = move_section_relative (ABCDEF, (-G,-H))
    (D0 == E0 == 0) or ({}[2])
    a,(p,q) = hyperbola_a_pq ( (A0,B0,C0,F0) )
    # Two foci:
    F1 = G+p,H+q ; F2 = G-p,H-q
    if flag :
        # Check calculations.
        # Produce hyperbola in standard form.
        ABCDEF_ = hyperbola_ABCDEF (2*a, F1, F2)
        # zip returns:
        # ( (ABCDEF[0], ABCDEF_[0]),
        #    .....................
        #   (ABCDEF[5], ABCDEF_[5]) )
        # (Both v,v_ must be 0) or (Both v,v_ must be non-zero.)
        for (v,v_) in zip(ABCDEF, ABCDEF_) : ( bool(v) == bool(v_) ) or ({}[3])
        set1 = set([ (v_/v) for (v_,v) in zip( ABCDEF_, (A,B,C,D,E,F) ) if v ])
        (len(set1) == 1) or ({}[4])
        
        if flag == 2 :
            print ()
            print (thisName)
            # Print results.
            str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'
            str2 = str1.format(A,B,C,D,E,F)
            print (str2)
            K, = set1
            if K != 1 :
                A_,B_,C_,D_,E_,F_ = ABCDEF_
                str3 = str1.format(A_,B_,C_,D_,E_,F_)
                print (str3, 'Standard form.')
            str4 = '({})x^2 + ({})y^2 + ({})xy + ({}) = 0'.format(A0,B0,C0,F0)
            print (str4, 'At origin.')
            
            F1x,F1y = F1 ; F2x,F2y = F2
            two_c = distance_between_foci = ( (F1x-F2x)**2 + (F1y-F2y)**2 ).sqrt()
            length_of_transverse_axis = 2*a
            e = distance_between_foci/length_of_transverse_axis
        
            for c in 'K G,H a 2*a two_c e'.split() :
                print ('   ', c, '=', eval(c))
            str5 = '({}, {})    (x - ({}))^2 + (y - ({}))^2 = 1'
            print ('    F1:', str5.format(F1x,F1y, F1x,F1y))
            print ('    F2:', str5.format(F2x,F2y, F2x,F2y))
            print ('    f1:', str5.format(p,q, p,q))
            print ('    f2:', str5.format(-p,-q, -p,-q))
            
            # Axis:
            Dx = F1x-F2x ; Dy = F1y - F2y
            # Dy x - Dx y + c = 0
            a1,b1 = Dy, -Dx
            # a1 x + b1 y + c1 = 0
            c1 = -(a1*F1x + b1*F1y)
            print ( '    axis: ({})x + ({})y + ({}) = 0'.format(a1,b1,c1) )
        
    return 2*a,F1,F2

Example
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Figure 1: Hyperbola with random position and orientation.


Equation of hyperbola is given.
This section calculates length of transverse axis,

A hyperbola has equation:

Calculate two foci and length of transverse axis.

# python code

ABCDEF = 176, -351, -336, -2528, 6504, -89104
two_a, F1, F2 = hyperbola_2a_F1_F2 (ABCDEF)
print ('Length of transverse axis =',two_a )
print ('F1 = ({},{})'.format(F1[0], F1[1] ))
print ('F2 = ({},{})'.format(F2[0], F2[1] ))
Length of transverse axis = 40
F1 = (35,-3)
F2 = (-13,11)

To check calculation and for more information about hyperbola:

# python code
hyperbola_2a_F1_F2 (ABCDEF, 2) # Set flag.
hyperbola_2a_F1_F2 (ABCDEF, flag = 2) :
(176.00)x^2 + (-351.00)y^2 + (-336)xy + (-2528.00)x + (6504.00)y + (-89104.0000) = 0
(-176.00)x^2 + (351.00)y^2 + (336.0)xy + (2528.000)x + (-6504.00)y + (89104.0000) = 0 Standard form.
(176.00)x^2 + (-351.00)y^2 + (-336)xy + (-90000.0000) = 0 At origin.
    K = -1
    G,H = (Decimal('11'), Decimal('4'))
    a = 20.0
    2*a = 40.0
    two_c = 50.0 
    e = 1.25
    F1: (35.0, -3)    (x - (35.0))^2 + (y - (-3))^2 = 1
    F2: (-13.0, 11)    (x - (-13.0))^2 + (y - (11))^2 = 1
    f1: (24.0, -7)    (x - (24.0))^2 + (y - (-7))^2 = 1
    f2: (-24.0, 7)    (x - (-24.0))^2 + (y - (7))^2 = 1
    axis: (-14)x + (-48.0)y + (346.0) = 0

Asymptotes of hyperbola

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Graph of hyperbola where .
Blue lines are asymptotes of hyperbola.
Hyperbola and asymptotes do not intersect.
Any line parallel to asymptote intersects hyperbola in exactly one place.

We are familiar with values where:

  • Length of transverse axis
  • Distance between foci

And we have said: Let What is ?

Value defines the In diagram, line segment is conjugate axis with length

Box at origin with length and width contains two special lines (blue lines) called


Let asymptote1 have equation

Hyperbola has equation

Calculate point of intersection of


From

Substitute for in

This does not make sense. There is no real point of intersection of


Let any line parallel to asymptote have equation:

Calculate point of intersection of


Provided that are non-zero, is a real number.


  • Asymptote is line whose position is as close as possible to hyperbola (both sides) without touching it.
  • Any line parallel to asymptote intersects hyperbola in exactly one place.

Line and hyperbola

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This section describes possibilities that arise when we consider intersection of line and hyperbola.

At origin
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Let hyperbola have equation

Let line have equation:

Let line intersect curve. For substitute

Expand simplify, gather like terms and result is quadratic function in

where:


From these results we can deduce:

  • If line is parallel to asymptote, c2 = 0.
  • If c2 == 0, line is parallel to asymptote.
  • If line is asymptote, both are 0.
  • If both are 0, line is asymptote.


These deductions are included in general case below.

Generally
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Let the conic section have the familiar equation:


Let a line have equation:


Let intersect the conic section. For in equation of conic section substitute



Expand simplify, gather like terms and result is quadratic function in


___ where:

_

_

_


If line is an asymptote, then _ _ in which case:

___ where:

___

and (roots of ) are the slopes of the 2 asymptotes.


Python code below recognizes whether or not line is asymptote or parallel to asymptote.

Implementation
[edit | edit source]
# python code
def hyperbola_and_line (ABCDEF, line_abc, flag = 0) :
    '''
This function calculates point/s of intersection (if any) of hyperbola and line.
hyperbola is: Ax^2 + By^2 + Cxy + Dx + Ey + F = 0
line is: ax + by + c = 0
To invoke:
[] = hyperbola_and_line (ABCDEF, line_abc[, flag]) 
[point1] = hyperbola_and_line (ABCDEF, line_abc[, flag]) 
[point1,point2] = hyperbola_and_line (ABCDEF, line_abc[, flag])
if line is asymptote or parallel to asymptote, output is type tuple.
if flag : check results.
if flag==2 : print results.
    '''
    
    (2 >= flag >= 0) or ({}[1])
    thisName = 'hyperbola_and_line (ABCDEF, line_abc, {}) :'.format(flag)
    a,b,c = [ dD(str(v)) for v in line_abc ]
    # a,b,c refer to line ax + by + c = 0.
    if a == b == 0 :
        print (thisName)
        print ('  At least one of a,b must be non-zero.')
        return None
    A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]

    output = []
    if b == 0:
        # Reverse x,y going in.
        result = hyperbola_and_line ( (B,A,C,E,D,F), (b,a,c))
        # Reverse x,y coming out.
        output = (type(result))([ v[::-1] for v in result ])
        # output is same type as result.
    else:
        # Axx + Byy + Cxy + Dx + Ey + F = 0
        # ax + by + c = 0 ; by = (- (ax + c))
        # Multiply equation of hyperbola by bb:
        # Abbxx + Bbyby + Cxbby + Dbbx + Ebby + Fbb = 0
        # For 'by' substitute '(- (ax + c))'.
        # Abbxx + B(-(ax+c))(-(ax+c)) + Cxb(-(ax+c)) + Dbbx + Eb(-(ax+c)) + Fbb = 0
        # Expand and result is quadratic function in x, (a_)xx + (b_)x + (c_) = 0 where
        a_ = A*b*b, + B*a*a, - C*a*b
        b_ = B*2*a*c, + D*b*b, - C*b*c, - E*b*a
        c_ = B*c*c, + F*b*b, - E*b*c
        a_,b_,c_ = [ sum_zero(v) for v in (a_,b_,c_) ]
        while 1:
            if a_ == 0 :
                # Line is parallel to asymptote.
                # values_of_x is of type tuple.
                if b_ == 0 :
                    # Line is asymptote. Return empty tuple.
                    values_of_x = tuple([]) ; break
                values_of_x = ( -c_/b_, ) ; break

            # values_of_x is of type list.
            two_a = 2*a_ ; discr = sum_zero((b_**2, - 4*a_*c_))
            if discr == 0 :
                # discr is 0 or very close to 0.
                values_of_x = [ -b_/two_a ] ; break

            if discr < 0 :
                values_of_x = [ ] ; break

            root = discr.sqrt()
            values_of_x = [ (-b_ - root)/two_a, (-b_ + root)/two_a ] ; break

        for x in values_of_x :
            by = -(a*x+c) ; y = by/b # Here is why b must be non-zero.
            output += [ (x,y) ]
        # output is same type as values_of_x.
        output = (type(values_of_x))(output)

    if flag :
        # Check results.
        errors = []
        for x,y in output :
            sum1 = sum_zero((A*x**2 , B*y**2 , C*x*y , D*x , E*y , F))
            if sum1 :
                errors += [ 'bad sum1: {} for point ({},{})'.format(sum1, x,y) ]
            sum2 = sum_zero((a*x , b*y , c))
            if sum2 :
                errors += [ 'bad sum2: {} for point ({},{})'.format(sum2, x,y) ]
        if errors or (flag == 2) :
            # Print results.
            print ()
            print (thisName)
            str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'
            print ( str1.format(A,B,C,D,E,F) )
            str3 = '({})x + ({})y + ({}) = 0'.format(a,b,c)
            print (str3)
            print ( 'ABCDEF = A,B,C,D,E,F = {}, {}, {}, {}, {}, {}'.format(A,B,C,D,E,F) )
            print ( 'abc = a,b,c = {}, {}, {}'.format(a,b,c) )
            
            str4 = 'output[{}]: ({},{}),    (x - ({}))^2 + (y - ({}))^2 = 1'
            for p in range (0,len(output)) :
                x,y = output[p]
                print ('   ', str4.format(p, x,y, x,y))
            for v in errors : print ('   ',v)
    # output may be empty: [] or (). () means asymptote.
    # or output may contain one point:
    #     [ (x1,y1) ] or ( (x1,y1), ). ( (x1,y1), ) means line is parallel to asymptote.
    # or output may contain two points: [ (x1,y1), (x2,y2) ]
    return output

Examples
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With no common point
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Diagram of hyperbola and two lines.
Both lines do not touch hyperbola.
Line 1 is asymptote.
Line 2 is not asymptote.

Line 1:

# python code

ABCDEF = A,B,C,D,E,F = 704, -1404, 1344, -11040, -41220, -161775

abc = a,b,c = 88.0, 234.0, 435.0
result = hyperbola_and_line (ABCDEF, abc) 
sx = 'result' ; print (sx, eval(sx))

Code recognizes that line is asymptote and returns empty tuple:

result ()

Line 2:

# python code

result = hyperbola_and_line (ABCDEF, (1,0,-10)) # x = 10.
sx = 'result' ; print (sx, eval(sx))

Code recognizes that line is not asymptote and returns empty list:

result []

With one common point
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Diagram of hyperbola and two lines.
Each line and hyperbola have one common point.
Line 1 (blue) is parallel to asymptote.
Line 2 (orange) is not parallel to asymptote.

Line 1:

# python code

ABCDEF = A,B,C,D,E,F = 704, -1404, 1344, -11040, -41220, -161775

abc = a,b,c = 88.0, 234.0, -1065
result = hyperbola_and_line (ABCDEF, abc) 
sx = 'result' ; print (sx, eval(sx))

Code recognizes that line is parallel to asymptote and returns tuple containing one point:

result ((24.6, -4.7),)

Line 2:

# python code

abc = a,b,c = -.96, -.28, 2.3
result = hyperbola_and_line (ABCDEF, abc) 
sx = 'result' ; print (sx, eval(sx))

Code recognizes that line is not parallel to asymptote and returns list containing one point:

result [ (5.4, -10.3) ]

With two common points
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Diagram of hyperbola and line.
Line and hyperbola have two common points.
When line and hyperbola have two common points, line cannot be parallel to asymptote.

Line 1:

# python code

ABCDEF = A,B,C,D,E,F = 704, -1404, 1344, -11040, -41220, -161775

abc = a,b,c = .96, .28, .2
result = hyperbola_and_line (ABCDEF, abc) 
sx = 'result' ; print (sx, eval(sx))

Code returns list containing two points:

result [ (1.425,-5.6), (4.575,-16.4) ]

Calculation of Asymptotes

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# python code

def asymptotes_of_hyperbola (ABCDEF, flag = 0) :
    '''
asymptote1, asymptote2 = asymptotes_of_hyperbola (ABCDEF [, flag])
Each asymptote is of form (a,b,c) where ax + by + c = 0.
if flag == 1: check results.
if flag == 2: check and print results.
    '''
    (2 >= flag >= 0) or ({}[3])
    thisName = 'asymptotes_of_hyperbola (ABCDEF, {}) :'.format(flag)

    A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]

    G,H = center_of_hyperbola (ABCDEF)
    while 1 :
        if A == B == 0 :
            if 0 in (C,F) :
                print (thisName)
                print ('    For rectangular hyperbola, both C,F must be non-zero.')
                return None
            asymptote1 = 1,0,-G # x = G
            asymptote2 = 0,1,-H # y = H
            output = [ asymptote1, asymptote2 ] ; break

        if B == 0 :
            # Reverse x,y going in.
            result = asymptotes_of_hyperbola ( (B,A,C,E,D,F), int(bool(flag)) )
            (a1,b1,c1),(a2,b2,c2) = result
            # Reverse x,y coming out.
            output = [ (b1,a1,c1), (b2,a2,c2) ] ; break

        _a,_b,_c = B,C,A
        discr = _b**2 - 4*_a*_c
        root = discr.sqrt()
        m1 = (-_b - root),(2*_a)
        m2 = (-_b + root),(2*_a)
        #     p
        # y = -x + c
        #     q
        # 
        # qy = px + c
        # px - qy + c = 0
        # ax + by + c = 0
        p,q = m1 ; a1,b1 = p,-q
        c1 = -(a1*G + b1*H)
        asymptote1 = a1,b1,c1
        p,q = m2 ; a2,b2 = p,-q
        c2 = -(a2*G + b2*H)
        asymptote2 = a2,b2,c2
        output = [ asymptote1, asymptote2 ]

        if flag :
            # Check results.
            values_of_c_ = []
            for a3,b3,c3 in output :
                # a3 x + b3 y + c3 = 0
                # b3 y = -(a3 x + c3)
                m = -a3/b3 ; c = -c3/b3
                b_ = sum_zero((2*B*c*m , c*C , D , E*m))
                b_ and ({}[6])
                values_of_c_ += [ E*c + F + B*c**2 ]
            c_1,c_2 = values_of_c_ # c_1,c_2 should be equal.
            sum_zero ((c_1, -c_2)) and ({}[6])
        break

    if flag :
        results = []
        for asymptote in output :
            result = hyperbola_and_line (ABCDEF, asymptote, 1)
            results += [ result ]
        set1 = { v==tuple([]) for v in results }
        error = (set1 != {True})
        if error or (flag == 2) :
            print (thisName)
            str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'
            a,b,c,d,e,f = [ float(v) for v in ABCDEF ]
            print ( '   ', str1.format(a,b,c,d,e,f) )
        
            asymptote1,asymptote2 = output
            a1,b1,c1 = [ float(v) for v in asymptote1 ]
            print ( '    asymptote1: ({})x + ({})y + ({}) = 0    {}'.format(a1,b1,c1, results[0]==()) )
            a2,b2,c2 = [ float(v) for v in asymptote2 ]
            print ( '    asymptote2: ({})x + ({})y + ({}) = 0    {}'.format(a2,b2,c2, results[1]==()) )
    return output

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Family of hyperbolas

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Family of hyperbolas

Asymptote1 (brown line):
Asymptote2 (green line):

Asymptotes of hyperbola are determined by coefficients

All hyperbolas in diagram have coefficients

Consider the family of hyperbolas that satisfy equation:

Provided that coefficients remain constant, hyperbolas of form are members of a family, all of which share the same asymptotes.


Examples of this family in diagram are:

Rectangular Hyperbolas

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Rectangular Hyperbolas are special cases of hyperbolas where the asymptotes are perpendicular. These have the general equation:

Diagram 1. Graph of rectangular hyperbola at origin.

A rectangular hyperbola has eccentricity

If hyperbola is expressed as then:

  • Distance from origin to focus:
  • Distance from origin to vertex:
  • Distance from origin to directrix:
  • because
  • Focus because
  • Focus
  • Vertex because
  • Vertex
  • Directrix 1 has equation: or
  • Directrix 2 has equation: or


It is obvious that asymptotes are the axes.

Details

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Diagram 2: Detail of quadrant 1 of Diagram 1.

Diagram 2 shows detail of quadrant 1 of Diagram 1.


  • Distance from origin to Directrix1 distance
  • Distance from origin to Vertex1 distance
  • Distance from origin to Focus1 distance


Line segment

  • is parallel to conjugate axis,
  • has length of conjugate axis:
  • has equation:


Directrix1 has equation:


Three circles are included for reference:

  • Circle1, green line through point has equation:
  • Circle2, magenta line through vertex has equation:
  • Circle3, orange line through focus has equation:

Examples

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Rectangular hyperbola at origin
[edit | edit source]
Figure 1: Rectangular hyperbola at origin with transverse axis horizontal.

Curve in Figure 1 has equation:


When distance distance

Vertex has coordinates Vertex has coordinates


Distance

Focus has coordinates Focus has coordinates


Distance

Point has coordinates Point has coordinates

Directrix1, directrix through point has equation

Directrix2, directrix through point has equation


Asymptote1 has equation:

Asymptote2 has equation:

Slope of asymptote1:

Slope of asymptote2:

Product of slopes: Asymptotes are perpendicular.

Rectangular hyperbola parallel to Y axis
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Figure 2: Rectangular hyperbola parallel to Y axis.

Curve is moved from origin to point

Curve in Figure 2 has equation:


Center of hyperbola, point has coordinates


Distance distance

Vertex has coordinates Vertex has coordinates


Distance distance

Focus has coordinates Focus has coordinates


Distance distance

Point has coordinates Point has coordinates

Directrix1, directrix through point has equation

Directrix2, directrix through point has equation



Asymptote1 has equation:

Asymptote2 has equation:

Slope of asymptote1:

Slope of asymptote2:

Product of slopes: Asymptotes are perpendicular.

Rectangular hyperbola, random
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Figure 3: Rectangular hyperbola with random position and orientation.

Curve is moved from origin to point and rotated.

In all rectangular hyperbolas:
* Point is mid-point of segment
* Point is mid-point of segment

Curve in Figure 3 has equation:


Center of hyperbola, point has coordinates


Distance distance

Vertex has coordinates Vertex has coordinates


Distance distance

Focus has coordinates Focus has coordinates


Distance distance

Point has coordinates Point has coordinates

Conjugate axis through point has equation

Directrix1, directrix through point has equation

Directrix2, directrix through point has equation


Asymptote1 has equation:

Asymptote2 has equation:

Slope of asymptote1:

Slope of asymptote2:

Product of slopes: Asymptotes are perpendicular.

Conic sections generally

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Within the two dimensional space of Cartesian Coordinate Geometry a conic section may be located anywhere and have any orientation.

This section examines the parabola, ellipse and hyperbola, showing how to calculate the equation of the conic section, and also how to calculate the foci and directrices given the equation.

Deriving the equation

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The curve is defined as a point whose distance to the focus and distance to a line, the directrix, have a fixed ratio, eccentricity Distance from focus to directrix must be non-zero.

Let the point have coordinates

Let the focus have coordinates

Let the directrix have equation where

Then

Square both sides:

Rearrange:

Expand simplify, gather like terms and result is:

where:

Note that values depend on:

  • non-zero. This method is not suitable for circle where
  • Sign of is not significant.
  • or and produce same result.

For example, directrix and directrix produce same result.

Implementation

[edit | edit source]
# python code
import decimal

dD = decimal.Decimal # Decimal object is like a float with (almost) unlimited precision.
dgt = decimal.getcontext()
Precision = dgt.prec = 22


def reduce_Decimal_number(number) :
# This function improves appearance of numbers.
# The technique used here is to perform the calculations using precision of 22,
# then convert to float or int to display result.
# -1e-22 becomes 0.
#  12.34999999999999999999 becomes 12.35
# -1.000000000000000000001 becomes -1.
# 1E+1 becomes 10.
# 0.3333333333333333333333 is unchanged.
#
    thisName = 'reduce_Decimal_number(number) :'
    if type(number) != dD : number = dD(str(number))

    f1 = float(number)
    if (f1 + 1) == 1 : return dD(0)
    if int(f1) == f1 : return dD(int(f1))
        
    dD1 = dD(str(f1))

    t1 = dD1.normalize().as_tuple()
    if (len(t1[1]) < 12) :
        # if number == 12.34999999999999999999, dD1 = 12.35
        return dD1

    return number


def ABCDEF_from_abc_epq (abc,epq,flag = 0) :
    '''
ABCDEF = ABCDEF_from_abc_epq (abc,epq[,flag]) 
    '''
    thisName = 'ABCDEF_from_abc_epq (abc,epq, {}) :'.format(bool(flag))
    a,b,c = [ dD(str(v)) for v in abc ]
    e,p,q = [ dD(str(v)) for v in epq ]

    divider = a**2 + b**2
    if divider == 0 :
        print (thisName, 'At least one of (a,b) must be non-zero.')
        return None
    if divider != 1 :
        root = divider.sqrt()
        a,b,c = [ (v/root) for v in (a,b,c) ]

    distance_from_focus_to_directrix = a*p + b*q + c
    if distance_from_focus_to_directrix == 0 :
        print (thisName, 'distance_from_focus_to_directrix must be non-zero.')
        return None

    X = e*e
    A = X*a**2 - 1
    B = X*b**2 - 1
    C = 2*X*a*b
    D = 2*p + 2*X*a*c
    E = 2*q + 2*X*b*c
    F = X*c**2 - p*p - q*q

    A,B,C,D,E,F = [ reduce_Decimal_number(v) for v in (A,B,C,D,E,F) ]

    if flag :
        print (thisName)
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(A,B,C,D,E,F)
        print (' ', str1)
        
    return (A,B,C,D,E,F)

Examples

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Parabola

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Every parabola has eccentricity

Quadratic function complies with definition of parabola.
Distance from point to focus
= distance from point to directrix = 10.
Distance from point to focus
= distance from point to directrix = 1.

Simple quadratic function:

Let focus be point

Let directrix have equation: or

# python code

p,q = 0,1
a,b,c = abc = 0,1,q
epq = 1,p,q

ABCDEF = ABCDEF_from_abc_epq (abc,epq,1)
print ('ABCDEF =', ABCDEF)
(-1)x^2 + (0)y^2 + (0)xy + (0)x + (4)y + (0) = 0
ABCDEF = (Decimal('-1'), Decimal('0'), Decimal('0'), Decimal('0'), Decimal('4'), Decimal('0'))

As conic section curve has equation:

Curve is quadratic function: or

For a quick check select some random points on the curve:

# python code

for x in (-2,4,6) :
    y = x**2/4
    print ('\nFrom point ({}, {}):'.format(x,y))
    distance_to_focus = ((x-p)**2 + (y-q)**2)**.5
    distance_to_directrix = a*x + b*y + c
    s1 = 'distance_to_focus' ; print (s1, eval(s1))
    s1 = 'distance_to_directrix' ; print (s1, eval(s1))
From point (-2, 1.0):
distance_to_focus 2.0
distance_to_directrix 2.0

From point (4, 4.0):
distance_to_focus 5.0
distance_to_directrix 5.0

From point (6, 9.0):
distance_to_focus 10.0
distance_to_directrix 10.0

[edit | edit source]

Curve in Figure 1 below has:

  • Directrix:
  • Focus:
  • Equation: or

Curve in Figure 2 below has:

  • Directrix:
  • Focus:
  • Equation: or

Curve in Figure 3 below has:

  • Directrix:
  • Focus:
  • Equation:

Ellipse

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Every ellipse has eccentricity

Ellipse with ecccentricity of 0.25 and center at origin.
Point1
Eccentricity
For every point on curve,

A simple ellipse:

Let focus be point where

Let directrix have equation: or

Let eccentricity

# python code
p,q = -1,0
e = 0.25
abc = a,b,c = 1,0,16
epq = e,p,q
ABCDEF_from_abc_epq  (abc,epq,1)
(-0.9375)x^2 + (-1)y^2 + (0)xy + (0)x + (0)y + (15) = 0

Ellipse has center at origin and equation:


Some basic checking:

# python code

points = (
    (-4  ,  0    ),
    (-3.5, -1.875),
    ( 3.5,  1.875),
    (-1  ,  3.75 ),
    ( 1  , -3.75 ),
)

A,B,F = -0.9375, -1, 15

for (x,y) in points :
    # Verify that point is on curve.
    (A*x**2 + B*y**2 + F) and 1/0 # Create exception if sum != 0.
    distance_to_focus = ( (x-p)**2 + (y-q)**2 )**.5
    distance_to_directrix = a*x + b*y + c
    e = distance_to_focus / distance_to_directrix
    s1 = 'x,y' ; print (s1, eval(s1))
    s1 = '    distance_to_focus, distance_to_directrix, e' ; print (s1, eval(s1))
x,y (-4, 0)
    distance_to_focus, distance_to_directrix, e (3.0, 12, 0.25)
x,y (-3.5, -1.875)
    distance_to_focus, distance_to_directrix, e (3.125, 12.5, 0.25)
x,y (3.5, 1.875)
    distance_to_focus, distance_to_directrix, e (4.875, 19.5, 0.25)
x,y (-1, 3.75)
    distance_to_focus, distance_to_directrix, e (3.75, 15.0, 0.25)
x,y (1, -3.75)
    distance_to_focus, distance_to_directrix, e (4.25, 17.0, 0.25)

Ellipses with ecccentricities from 0.1 to 0.9.
As eccentricity approaches shape of ellipse approaches shape of circle.
As eccentricity approaches shape of ellipse approaches shape of parabola.

The effect of eccentricity.


All ellipses in diagram have:

  • Focus at point
  • Directrix with equation


Five ellipses are shown with eccentricities varying from to

[edit | edit source]

Curve in Figure 1 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Curve in Figure 2 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Curve in Figure 3 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Hyperbola

[edit | edit source]

Every hyperbola has eccentricity

Hyperbola with eccentricity of 1.5 and center at origin.
Point1
Eccentricity
For every point on curve,

A simple hyperbola:

Let focus be point where

Let directrix have equation: or

Let eccentricity

# python code
p,q = 0,-9
e = 1.5
abc = a,b,c = 0,1,4
epq = e,p,q
ABCDEF_from_abc_epq  (abc,epq,1)
(-1)x^2 + (1.25)y^2 + (0)xy + (0)x + (0)y + (-45) = 0

Hyperbola has center at origin and equation:

Some basic checking:

# python code

four_points = pt1,pt2,pt3,pt4 = (-7.5,9),(-7.5,-9),(22.5,21),(22.5,-21)
for (x,y) in four_points :
    # Verify that point is on curve.
    sum = 1.25*y**2 - x**2 - 45
    sum and 1/0 # Create exception if sum != 0.
    distance_to_focus = ( (x-p)**2 + (y-q)**2 )**.5
    distance_to_directrix = a*x + b*y + c
    e = distance_to_focus / distance_to_directrix
    s1 = 'x,y' ; print (s1, eval(s1))
    s1 = '    distance_to_focus, distance_to_directrix, e' ; print (s1, eval(s1))
x,y (-7.5, 9)
    distance_to_focus, distance_to_directrix, e (19.5, 13.0, 1.5)
x,y (-7.5, -9)
    distance_to_focus, distance_to_directrix, e (7.5, -5.0, -1.5)
x,y (22.5, 21)
    distance_to_focus, distance_to_directrix, e (37.5, 25.0, 1.5)
x,y (22.5, -21)
    distance_to_focus, distance_to_directrix, e (25.5, -17.0, -1.5)

Hyperbolas with ecccentricities from 1.5 to 20.
As eccentricity increases, curve approaches directrix:
As eccentricity approaches shape of curve approaches shape of parabola.

The effect of eccentricity.


All hyperbolas in diagram have:

  • Focus at point
  • Directrix with equation


Six hyperbolas are shown with eccentricities varying from to

[edit | edit source]

Curve in Figure 1 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Curve in Figure 2 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Curve in Figure 3 below has:

  • Directrix:
  • Focus:
  • Eccentricity:
  • Equation:

Reversing the process

[edit | edit source]

The expression "reversing the process" means calculating the values of focus and directrix when given the equation of the conic section, the familiar values

Consider the equation of a simple ellipse: This is a conic section where

This ellipse may be expressed as a format more appealing to the eye than numbers containing fractions or decimals.

However, when this ellipse is expressed as this format is the ellipse expressed in "standard form," a notation that greatly simplifies the calculation of

Modify the equations for slightly:

or

or

In substitute for

is a quadratic equation in where:

Because is a quadratic equation, the solution of may contain an unwanted value of that will be eliminated later.

From and

Because

Implementation

[edit | edit source]
# python code


def solve_quadratic (abc) :
    '''
result = solve_quadratic (abc)
result may be :
    []
    [ root1 ]
    [ root1, root2 ]
'''
    a,b,c = abc

    if a == 0 : return [ -c/b ]
    
    disc = b**2 - 4*a*c
    
    if disc < 0 : return []

    two_a = 2*a
    if disc == 0 : return [ -b/two_a ]

    root = disc.sqrt()
    r1,r2 = (-b - root)/two_a, (-b + root)/two_a
    return [r1,r2]


def calculate_Kab (ABC, flag=0) :
    '''
result = calculate_Kab (ABC)
result may be :
    []
    [tuple1]
    [tuple1,tuple2]
'''
    thisName = 'calculate_Kab (ABC, {}) :'.format(bool(flag))
    A_,B_,C_ = [ dD(str(v)) for v in ABC ]
    # Quadratic function in K: (a_)K**2 + (b_)K + (c_) = 0
    a_ = 4*A_*B_ - C_*C_
    b_ = 4*(A_+B_)
    c_ = 4

    values_of_K = solve_quadratic ((a_,b_,c_))

    if flag :
        print (thisName)
        str1 = '  A_,B_,C_' ; print (str1,eval(str1))
        str1 = '  a_,b_,c_' ; print (str1,eval(str1))
        print ('  y = ({})x^2 + ({})x + ({})'.format( float(a_), float(b_), float(c_) ))
        str1 = '  values_of_K' ; print (str1,eval(str1))

    output = []
    for K in values_of_K :
        A,B,C = [ reduce_Decimal_number(v*K) for v in (A_,B_,C_) ]
        X = A + B + 2
        if X <= 0 :
            # Here is one place where the spurious value of K may be eliminated.
            if flag : print ('  K = {}, X = {}, continuing.'.format(K, X))
            continue

        aa = reduce_Decimal_number((A + 1)/X)

        if flag :
            print ('  K =', K)
            for strx in ('A', 'B', 'C', 'X', 'aa') :
                print ('   ', strx, eval(strx))

        if aa == 0 :
            a = dD(0) ; b = dD(1)
        else :
            a = aa.sqrt() ; b = C/(2*X*a)
        
        Kab = [ reduce_Decimal_number(v) for v in (K,a,b) ]
        output += [ Kab ]
    if flag:
        print (thisName)
        for t in range (0, len(output)) :
            str1 = '  output[{}] = {}'.format(t,output[t])
            print (str1)
    return output

More calculations

[edit | edit source]

The values

In replace

Expand simplify, gather like terms and result is quadratic function in

where:

Therefore:

For parabola, there is one value of because there is one directrix.

For ellipse and hyperbola, there are two values of because there are two directrices.

Implementation

[edit | edit source]
# python code

def compare_ABCDEF1_ABCDEF2 (ABCDEF1, ABCDEF2) :
    '''
status = compare_ABCDEF1_ABCDEF2 (ABCDEF1, ABCDEF2)

This function compares the two conic sections.
"0.75x^2 + y^2 + 3 = 0" and "3x^2 + 4y^2 + 12 = 0" compare as equal.
"0.75x^2 + y^2 + 3 = 0" and "3x^2 + 4y^2 + 10 = 0" compare as not equal.

(0.24304)x^2 + (1.49296)y^2 + (-4.28544)xy + (159.3152)x + (-85.1136)y + (2858.944) = 0
            and
(-0.0784)x^2 + (-0.4816)y^2 + (1.3824)xy + (-51.392)x + (27.456)y + (-922.24) = 0
            are verified as the same curve.

>>> abcdef1 = (0.24304, 1.49296, -4.28544, 159.3152, -85.1136, 2858.944)
>>> abcdef2 = (-0.0784, -0.4816, 1.3824, -51.392, 27.456, -922.24)
>>> [ (v[0]/v[1]) for v in zip(abcdef1, abcdef2) ]
[-3.1, -3.1, -3.1, -3.1, -3.1, -3.1]
set ([-3.1, -3.1, -3.1, -3.1, -3.1, -3.1]) = {-3.1}
'''
    thisName = 'compare_ABCDEF1_ABCDEF2 (ABCDEF1, ABCDEF2) :'

    # For each value in ABCDEF1, ABCDEF2, both value1 and value2 must be 0
    # or both value1 and value2 must be non-zero.
    for v1,v2 in zip (ABCDEF1, ABCDEF2) :
        status = (bool(v1) == bool(v2))
        if not status :
            print (thisName)
            print ('  mismatch:',v1,v2)
            return status

    # Results of v1/v2 must all be the same.
    set1 = { (v1/v2) for (v1,v2) in zip (ABCDEF1, ABCDEF2) if v2 }
    status = (len(set1) == 1)
    if status : quotient, = list(set1)
    else : quotient = '??'
    
    L1 = [] ; L2 = [] ; L3 = []

    for m in range (0,6) :
        bottom = ABCDEF2[m]
        if not bottom : continue
        top = ABCDEF1[m]
        L1 += [ str(top) ] ; L3 += [ str(bottom) ]
    for m in range (0,len(L1)) :
        L2 += [ (sorted( [ len(v) for v in (L1[m], L3[m]) ] ))[-1] ] # maximum value.
    for m in range (0,len(L1)) :
        max = L2[m]
        L1[m] = ( (' '*max)+L1[m] )[-max:] # string right justified.
        L2[m] = ( '-'*max )
        L3[m] = ( (' '*max)+L3[m] )[-max:] # string right justified.
    print ('   ', '   '.join(L1))
    print ('   ', ' = '.join(L2), '=', quotient)
    print ('   ', '   '.join(L3))
    return status


def calculate_abc_epq (ABCDEF_, flag = 0) :
    '''
result = calculate_abc_epq (ABCDEF_ [, flag])
For parabola, result is:
    [((a,b,c), (e,p,q))]
For ellipse or hyperbola, result is:
    [((a1,b1,c1), (e,p1,q1)), ((a2,b2,c2), (e,p2,q2))]
'''

    thisName = 'calculate_abc_epq (ABCDEF, {}) :'.format(bool(flag))

    ABCDEF = [ dD(str(v)) for v in ABCDEF_ ]
    if flag :
        v1,v2,v3,v4,v5,v6 = ABCDEF
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(v1,v2,v3,v4,v5,v6)
        print('\n' + thisName, 'enter')
        print(str1)

    result = calculate_Kab (ABCDEF[:3], flag)

    output = []
    for (K,a,b) in result :
        A,B,C,D,E,F = [ reduce_Decimal_number(K*v) for v in ABCDEF ]

        X = A + B + 2
        e = X.sqrt()
        # Quadratic function in c: (a_)c**2 + (b_)c + (c_) = 0
        # Directrix has equation: ax + by + c = 0.
        a_ = 4*X*( 1 - X  )
        b_ = 4*X*( D*a + E*b )
        c_ = -D*D - E*E - 4*F
        
        values_of_c = solve_quadratic((a_,b_,c_))
        # values_of_c may be empty in which case this value of K is not used.

        for c in values_of_c :
            p = (D - 2*X*a*c)/2
            q = (E - 2*X*b*c)/2
            abc = [ reduce_Decimal_number(v) for v in (a,b,c) ]
            epq = [ reduce_Decimal_number(v) for v in (e,p,q) ]
            output += [ (abc,epq) ]
        if flag :
            print (thisName)
            str1 = '  ({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(A,B,C,D,E,F)
            print (str1)
            if values_of_c : str1 = '  K = {}. values_of_c = {}'.format(K, values_of_c)
            else : str1 = '  K = {}. values_of_c = {}'.format(K, 'EMPTY')
            print (str1)
    if len(output) not in (1,2) :
        # This should be impossible.
        print (thisName)
        print ('  Internal error: len(output) =', len(output))
        1/0

    if flag :
        # Check output and print results.

        L1 = []
        for ((a,b,c),(e,p,q)) in output :
            print ('  e =',e)
            print ('    directrix: ({})x + ({})y + ({}) = 0'.format(a,b,c) )
            print ('    for focus : p, q = {}, {}'.format(p,q))
            # A small circle at focus for grapher.
            print ('    (x - ({}))^2 + (y - ({}))^2 = 1'.format(p,q))

            # normal through focus :
            a_,b_ = b,-a
            # normal through focus : a_ x + b_ y + c_ = 0
            c_ = reduce_Decimal_number(-(a_*p + b_*q))
            print ('    normal through focus: ({})x + ({})y + ({}) = 0'.format(a_,b_,c_) )
            L1 += [ (a_,b_,c_) ]

            _ABCDEF = ABCDEF_from_abc_epq ((a,b,c),(e,p,q))
            # This line checks that values _ABCDEF, ABCDEF make sense when compared against each other.
            if not compare_ABCDEF1_ABCDEF2 (_ABCDEF, ABCDEF) :
                print ('    _ABCDEF =',_ABCDEF)
                print ('    ABCDEF =',ABCDEF)
                2/0
        # This piece of code checks that normal through one focus is same as normal through other focus.
        # Both of these normals, if there are 2, should be same line.
        # It also checks that 2 directrices, if there are 2, are parallel.
        set2 = set(L1)
        if len(set2) != 1 :
            print ('    set2 =',set2)
            3/0

    return output

Examples

[edit | edit source]

Parabola

[edit | edit source]
Graph of parabola
Equation of parabola is given.
This section calculates

Given equation of conic section:

Calculate

# python code

input = ( 16, 9, -24, 410, -420, 3175 )

(abc,epq), = calculate_abc_epq  (input)

s1 = 'abc' ; print (s1, eval(s1))
s1 = 'epq' ; print (s1, eval(s1))
abc [Decimal('0.6'), Decimal('0.8'), Decimal('3')]
epq [Decimal('1'), Decimal('-10'), Decimal('6')]

interpreted as:

Directrix:

Eccentricity:

Focus:

Because eccentricity is curve is parabola.

Because curve is parabola, there is one directrix and one focus.

For more insight into the method of calculation and also to check the calculation:

calculate_abc_epq  (input, 1) # Set flag to 1.
calculate_abc_epq (ABCDEF, True) : enter
(16)x^2 + (9)y^2 + (-24)xy + (410)x + (-420)y + (3175) = 0 # This equation of parabola is not in standard form.
calculate_Kab (ABC, True) :
  A_,B_,C_ (Decimal('16'), Decimal('9'), Decimal('-24'))
  a_,b_,c_ (Decimal('0'), Decimal('100'), 4)
  y = (0.0)x^2 + (100.0)x + (4.0)
  values_of_K [Decimal('-0.04')]
  K = -0.04
    A -0.64
    B -0.36
    C 0.96
    X 1.00
    aa 0.36
calculate_Kab (ABC, True) :
  output[0] = [Decimal('-0.04'), Decimal('0.6'), Decimal('0.8')]
calculate_abc_epq (ABCDEF, True) :
  (-0.64)x^2 + (-0.36)y^2 + (0.96)xy + (-16.4)x + (16.8)y + (-127) = 0 # This is equation of parabola in standard form.
  K = -0.04. values_of_c = [Decimal('3')]
  e = 1
    directrix: (0.6)x + (0.8)y + (3) = 0
    for focus : p, q = -10, 6
    (x - (-10))^2 + (y - (6))^2 = 1
    normal through focus: (0.8)x + (-0.6)y + (11.6) = 0
    # This is proof that equation supplied and equation in standard form are same curve.
    -0.64   -0.36   0.96   -16.4   16.8   -127
    ----- = ----- = ---- = ----- = ---- = ---- = -0.04 # K
       16       9    -24     410   -420   3175

Ellipse

[edit | edit source]
Graph of ellipse
Equation of ellipse is given.
This section calculates

Given equation of conic section:

Calculate

# python code

input = ( 481, 369, -384, 5190, 5670, 7650 )

(abc1,epq1),(abc2,epq2) = calculate_abc_epq  (input)

s1 = 'abc1' ; print (s1, eval(s1))
s1 = 'epq1' ; print (s1, eval(s1))
s1 = 'abc2' ; print (s1, eval(s1))
s1 = 'epq2' ; print (s1, eval(s1))
abc1 [Decimal('0.6'), Decimal('0.8'), Decimal('-3')]
epq1 [Decimal('0.8'), Decimal('-3'), Decimal('-3')]
abc2 [Decimal('0.6'), Decimal('0.8'), Decimal('37')]
epq2 [Decimal('0.8'), Decimal('-18.36'), Decimal('-23.48')]

interpreted as:

Directrix 1:

Eccentricity:

Focus 1:

Directrix 2:

Eccentricity:

Focus 2:

Because eccentricity is curve is ellipse.

Because curve is ellipse, there are two directrices and two foci.

For more insight into the method of calculation and also to check the calculation:

calculate_abc_epq  (input, 1) # Set flag to 1.
calculate_abc_epq (ABCDEF, True) : enter
(481)x^2 + (369)y^2 + (-384)xy + (5190)x + (5670)y + (7650) = 0 # Not in standard form.
calculate_Kab (ABC, True) :
  A_,B_,C_ (Decimal('481'), Decimal('369'), Decimal('-384'))
  a_,b_,c_ (Decimal('562500'), Decimal('3400'), 4)
  y = (562500.0)x^2 + (3400.0)x + (4.0)
  values_of_K [Decimal('-0.004444444444444444444444'), Decimal('-0.0016')]
  # Unwanted value of K is rejected here.
  K = -0.004444444444444444444444, X = -1.777777777777777777778, continuing.
  K = -0.0016
    A -0.7696
    B -0.5904
    C 0.6144
    X 0.6400
    aa 0.36
calculate_Kab (ABC, True) :
  output[0] = [Decimal('-0.0016'), Decimal('0.6'), Decimal('0.8')]
calculate_abc_epq (ABCDEF, True) :
  # Equation of ellipse in standard form.
  (-0.7696)x^2 + (-0.5904)y^2 + (0.6144)xy + (-8.304)x + (-9.072)y + (-12.24) = 0
  K = -0.0016. values_of_c = [Decimal('-3'), Decimal('37')]
  e = 0.8
    directrix: (0.6)x + (0.8)y + (-3) = 0
    for focus : p, q = -3, -3
    (x - (-3))^2 + (y - (-3))^2 = 1
    normal through focus: (0.8)x + (-0.6)y + (0.6) = 0
    # Method calculates equation of ellipse using these values of directrix, eccentricity and focus.
    # Method then verifies that calculated and supplied values are the same curve.
    -0.7696   -0.5904   0.6144   -8.304   -9.072   -12.24
    ------- = ------- = ------ = ------ = ------ = ------ = -0.0016 # K
        481       369     -384     5190     5670     7650
  e = 0.8
    directrix: (0.6)x + (0.8)y + (37) = 0
    for focus : p, q = -18.36, -23.48
    (x - (-18.36))^2 + (y - (-23.48))^2 = 1
    normal through focus: (0.8)x + (-0.6)y + (0.6) = 0 # Same as normal above.
    # Method calculates equation of ellipse using these values of directrix, eccentricity and focus.
    # Method then verifies that calculated and supplied values are the same curve.
    -0.7696   -0.5904   0.6144   -8.304   -9.072   -12.24
    ------- = ------- = ------ = ------ = ------ = ------ = -0.0016 # K
        481       369     -384     5190     5670     7650

Hyperbola

[edit | edit source]
Graph of hyperbola
Equation of hyperbola is given.
This section calculates

Given equation of conic section:

Calculate

# python code

input = ( 7, 0, -24, 90, 216, -81 )

(abc1,epq1),(abc2,epq2) = calculate_abc_epq  (input)

s1 = 'abc1' ; print (s1, eval(s1))
s1 = 'epq1' ; print (s1, eval(s1))
s1 = 'abc2' ; print (s1, eval(s1))
s1 = 'epq2' ; print (s1, eval(s1))
abc1 [Decimal('0.6'), Decimal('0.8'), Decimal('-3')]
epq1 [Decimal('1.25'), Decimal('0'), Decimal('-3')]
abc2 [Decimal('0.6'), Decimal('0.8'), Decimal('-22.2')]
epq2 [Decimal('1.25'), Decimal('18'), Decimal('21')]

interpreted as:

Directrix 1:

Eccentricity:

Focus 1:

Directrix 2:

Eccentricity:

Focus 2:

Because eccentricity is curve is hyperbola.

Because curve is hyperbola, there are two directrices and two foci.

For more insight into the method of calculation and also to check the calculation:

calculate_abc_epq  (input, 1) # Set flag to 1.
calculate_abc_epq (ABCDEF, True) : enter
# Given equation is not in standard form.
(7)x^2 + (0)y^2 + (-24)xy + (90)x + (216)y + (-81) = 0
calculate_Kab (ABC, True) :
  A_,B_,C_ (Decimal('7'), Decimal('0'), Decimal('-24'))
  a_,b_,c_ (Decimal('-576'), Decimal('28'), 4)
  y = (-576.0)x^2 + (28.0)x + (4.0)
  values_of_K [Decimal('0.1111111111111111111111'), Decimal('-0.0625')]
  K = 0.1111111111111111111111
    A 0.7777777777777777777777
    B 0
    C -2.666666666666666666666
    X 2.777777777777777777778
    aa 0.64
  K = -0.0625
    A -0.4375
    B 0
    C 1.5
    X 1.5625
    aa 0.36
calculate_Kab (ABC, True) :
  output[0] = [Decimal('0.1111111111111111111111'), Decimal('0.8'), Decimal('-0.6')]
  output[1] = [Decimal('-0.0625'), Decimal('0.6'), Decimal('0.8')]
calculate_abc_epq (ABCDEF, True) :
  # Here is where unwanted value of K is rejected.
  (0.7777777777777777777777)x^2 + (0)y^2 + (-2.666666666666666666666)xy + (10)x + (24)y + (-9) = 0
  K = 0.1111111111111111111111. values_of_c = EMPTY
calculate_abc_epq (ABCDEF, True) :
  # Equation of hyperbola in standard form.
  (-0.4375)x^2 + (0)y^2 + (1.5)xy + (-5.625)x + (-13.5)y + (5.0625) = 0
  K = -0.0625. values_of_c = [Decimal('-3'), Decimal('-22.2')]
  e = 1.25
    directrix: (0.6)x + (0.8)y + (-3) = 0
    for focus : p, q = 0, -3
    (x - (0))^2 + (y - (-3))^2 = 1
    normal through focus: (0.8)x + (-0.6)y + (-1.8) = 0
    # Method calculates equation of hyperbola using these values of directrix, eccentricity and focus.
    # Method then verifies that calculated and given values are the same curve.
    -0.4375   1.5   -5.625   -13.5   5.0625
    ------- = --- = ------ = ----- = ------ = -0.0625 # K
          7   -24       90     216      -81
  e = 1.25
    directrix: (0.6)x + (0.8)y + (-22.2) = 0
    for focus : p, q = 18, 21
    (x - (18))^2 + (y - (21))^2 = 1
    normal through focus: (0.8)x + (-0.6)y + (-1.8) = 0 # Same as normal above.
    # Method calculates equation of hyperbola using these values of directrix, eccentricity and focus.
    # Method then verifies that calculated and given values are the same curve.
    -0.4375   1.5   -5.625   -13.5   5.0625
    ------- = --- = ------ = ----- = ------ = -0.0625 # K
          7   -24       90     216      -81

Slope of curve

[edit | edit source]

Given equation of conic section:

differentiate both sides with respect to




For slope horizontal:

For slope vertical:

For given slope

Implementation

[edit | edit source]
# python code

def three_slopes (ABCDEF, slope, flag = 0) :
    '''
equation1, equation2, equation3 = three_slopes (ABCDEF, slope[, flag]) 
equation1 is equation for slope horizontal.
equation2 is equation for slope vertical.
equation3 is equation for slope supplied.
All equations are in format (a,b,c) where ax + by + c = 0.
    '''
    A,B,C,D,E,F = ABCDEF
    output = []

    abc = 2*A, C, D ; output += [ abc ]
    abc = C, 2*B, E ; output += [ abc ]

    m = slope
    # m(Cx + 2By + E) = -2Ax - Cy - D
    # mCx + m2By + mE = -2Ax - Cy - D
    # mCx + 2Ax + m2By + Cy + mE + D = 0
    abc = m*C + 2*A, m*2*B + C, m*E + D ; output += [ abc ]

    if flag :
        str1 = '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format (A,B,C,D,E,F)
        print (str1)
        a,b,c = output[0]
        str1 = 'For slope horizontal: ({})x + ({})y + ({}) = 0'.format (a,b,c)
        print (str1)
        a,b,c = output[1]
        str1 = 'For slope vertical: ({})x + ({})y + ({}) = 0'.format (a,b,c)
        print (str1)
        a,b,c = output[2]
        str1 = 'For slope {}: ({})x + ({})y + ({}) = 0'.format (slope, a,b,c)
        print (str1)
        
    return output

Examples

[edit | edit source]

Quadratic function

[edit | edit source]
y = f(x)
[edit | edit source]
Graph of quadratic function
At intersection of and curve, slope = .
At intersection of and curve, slope = .
Slope of curve is never vertical.

Consider conic section:

This is quadratic function:

Slope of this curve:

Produce values for slope horizontal, slope vertical and slope

# python code

ABCDEF = A,B,C,D,E,F = -1,0,0,14,4,39 # quadratic 
three_slopes (ABCDEF, 5, 1)
(-1)x^2 + (0)y^2 + (0)xy + (14)x + (4)y + (39) = 0
For slope horizontal: (-2)x + (0)y + (14) = 0  # x = 7
For slope vertical: (0)x + (0)y + (4) = 0      # This does not make sense. 
                                               # Slope is never vertical. 
For slope 5: (-2)x + (0)y + (34) = 0           # x = 17.

Check results:

# python code

for x in (7,17) :
    m = (2*x - 14)/4
    s1 = 'x,m' ; print (s1, eval(s1))
x,m (7, 0.0)   # When x = 7, slope = 0. 
x,m (17, 5.0)  # When x = 17, slope = 5.

x = f(y)
[edit | edit source]
Graph of quadratic function
At intersection of and curve, slope is vertical.
At intersection of and curve, slope = .
Slope of curve is never horizontal.

Consider conic section:

This is quadratic function:

Slope of this curve:

Produce values for slope horizontal, slope vertical and slope

# python code

ABCDEF = A,B,C,D,E,F = 0,-1,0,-4,-14,-5 # quadratic x = f(y)
three_slopes (ABCDEF, 0.5, 1)
(0)x^2 + (-1)y^2 + (0)xy + (-4)x + (-14)y + (-5) = 0
For slope horizontal: (0)x + (0)y + (-4) = 0    # This does not make sense.
                                                # Slope is never horizontal.
For slope vertical: (0)x + (-2)y + (-14) = 0    # y = -7
For slope 0.5: (0.0)x + (-1.0)y + (-11.0) = 0   # y = -11

Check results:

# python code

for y in (-7,-11) :
    top = -4 ; bottom = 2*y + 14
    if bottom == 0 :
        print ('y,m',y,'{}/{}'.format(top,bottom))
        continue
    m = top/bottom
    s1 = 'y,m' ; print (s1, eval(s1))
y,m -7 -4/0     # When y = -7, slope is vertical.
y,m (-11, 0.5)  # When y = -11, slope is 0.5.

Parabola

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Graph of parabola
At intersection of and curve, slope is horizontal.
At intersection of and curve, slope is vertical.
At intersection of and curve, slope = .
Slope of curve is never because axis has slope and curve is never parallel to axis.

Consider conic section:

This curve is a parabola.


Produce values for slope horizontal, slope vertical and slope

# python code

ABCDEF = A,B,C,D,E,F = 9,16,-24,104,28,-144 # parabola
three_slopes (ABCDEF, 2, 1)
(9)x^2 + (16)y^2 + (-24)xy + (104)x + (28)y + (-144) = 0
For slope horizontal: (18)x + (-24)y + (104) = 0
For slope vertical: (-24)x + (32)y + (28) = 0
For slope 2: (-30)x + (40)y + (160) = 0

Because all 3 lines are parallel to axis, all 3 lines have slope


Produce values for slope horizontal, slope vertical and slope

# python code

three_slopes (ABCDEF, 0.75, 1)
(9)x^2 + (16)y^2 + (-24)xy + (104)x + (28)y + (-144) = 0
For slope horizontal: (18)x + (-24)y + (104) = 0 # Same as above.
For slope vertical: (-24)x + (32)y + (28) = 0    # Same as above.
For slope 0.75: (0.0)x + (0.0)y + (125.0) = 0    # Impossible.

Axis has slope and curve is never parallel to axis.

Ellipse

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Graph of ellipse
At intersection of and curve, slope is horizontal.
At intersection of and curve, slope is vertical.
At intersection of and curve, slope =

Consider conic section:

This curve is an ellipse.


Produce values for slope horizontal, slope vertical and slope

# python code

ABCDEF = A,B,C,D,E,F = 1771, 1204, 1944, -44860, -18520, 214400 # ellipse
three_slopes (ABCDEF, -1, 1)
(1771)x^2 + (1204)y^2 + (1944)xy + (-44860)x + (-18520)y + (214400) = 0
For slope horizontal: (3542)x + (1944)y + (-44860) = 0
For slope vertical: (1944)x + (2408)y + (-18520) = 0
For slope -1: (1598)x + (-464)y + (-26340) = 0

Because curve is closed loop, slope of curve may be any value including

If slope of curve is given as it means that curve is vertical at that point and tangent to curve has equation

For any given slope there are always 2 points on opposite sides of curve where tangent to curve at each of those points has the given slope.

Hyperbola

[edit | edit source]
Graph of hyperbola
At intersection of and curve, slope is horizontal.
and curve do not intersect. Slope is never vertical.
At intersection of and curve, slope =

Consider conic section:

This curve is a hyperbola.


Produce values for slope horizontal, slope vertical and slope

# python code

ABCDEF = A,B,C,D,E,F = -351, 176, -336, 4182, -3824, -16231 # hyperbola
three_slopes (ABCDEF, 2, 1)
(-351)x^2 + (176)y^2 + (-336)xy + (4182)x + (-3824)y + (-16231) = 0
For slope horizontal: (-702)x + (-336)y + (4182) = 0
For slope vertical: (-336)x + (352)y + (-3824) = 0
For slope 2: (-1374)x + (368)y + (-3466) = 0

Latera recta et cetera

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"Latus rectum" is a Latin expression meaning "straight side." According to Google, the Latin plural of "latus rectum" is "latera recta," but English allows "latus rectums" or possibly "lati rectums." The title of this section is poetry to the eyes and music to the ears of a Latin student and this author hopes that the gentle reader will permit such poetic licence in a mathematical topic.

The translation of the title is "Latus rectums and other things." This section describes the calculation of interesting items associated with the ellipse: latus rectums, major axis, minor axis, focal chords, directrices and various points on these lines.


When given the equation of an ellipse, the first thing is to calculate eccentricity, foci and directrices as shown above. Then verify that the curve is in fact an ellipse.

From these values everything about the ellipse may be calculated. For example:

Graph of ellipse

Axis : (-0.8)x + (-0.6)y + (9.4) = 0
Eccentricity = 0.9

Directrix 2 : (0.6)x + (-0.8)y + (2) = 0
Latus rectum RS : (0.6)x + (-0.8)y + (-0.8) = 0
Minor axis : (0.6)x + (-0.8)y + (-12.73684210526315789474) = 0
Latus rectum PU : (0.6)x + (-0.8)y + (-24.67368421052631578947) = 0
Directrix 1 : (0.6)x + (-0.8)y + (-27.47368421052631578947) = 0

= (6.32, 7.24)
= (7.204210526315789473684, 6.061052631578947368421)
F2 = (8, 5)
M = (15.16210526315789473684, -4.54947368421052631579)
F1 = (22.32421052631578947368, -14.09894736842105263158)
= (23.12, -15.16)
= (24.00421052631578947368, -16.33894736842105263158)

P = (20.30821052631578947368, -15.61094736842105263158)
Q = (10.53708406832736953616, -8.018239580333420216299)
R = (5.984, 3.488)
S = (10.016, 6.512)
T = (19.78712645798841993752, -1.080707788087632415281)
U = (24.34021052631578947368, -12.58694736842105263158)

Distance between directrices: = 29.47368421052631578947
Length of major axis: = 26.52631578947368421052
Distance between foci: = 23.87368421052631578947
Length of minor axis: QT = 11.56255298707631300170
Length of latus rectum: RS = PU = 5.04

Consider conic section:

This curve is ellipse with random orientation.

# python code

ABCDEF = A,B,C,D,E,F = 1771, 1204, 1944, -44860, -18520, 214400 # ellipse

result = calculate_abc_epq(ABCDEF)
(len(result) == 2) or 1/0

# ellipse or hyperbola
(abc1,epq1), (abc2,epq2) = result
a1,b1,c1 = abc1 ; e1,p1,q1 = epq1
a2,b2,c2 = abc2 ; e2,p2,q2 = epq2
(e1 == e2) or 2/0
(1 > e1 > 0) or 3/0

print ( '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(A,B,C,D,E,F) )
A,B,C,D,E,F = ABCDEF_from_abc_epq(abc1,epq1)
print ('Equation of ellipse in standard form:')
print ( '({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(A,B,C,D,E,F) )
(1771)x^2 + (1204)y^2 + (1944)xy + (-44860)x + (-18520)y + (214400) = 0
Equation of ellipse in standard form:
(-0.7084)x^2 + (-0.4816)y^2 + (-0.7776)xy + (17.944)x + (7.408)y + (-85.76) = 0
# python code
def sum_zero(input) :
    '''
sum = sum_zero(input)
If sum is close to 0 and Tolerance permits, sum is returned as 0.
For example: if input contains (2, -1.999999999999999999999)
this function returns sum of these 2 values as 0.
    '''
    global Tolerance
    sump = sumn = 0
    for v in input :
        if v > 0 : sump += v
        elif v < 0 : sumn -= v
    sum = sump - sumn
    if abs(sum) < Tolerance : return (type(Tolerance))(0)
    min, max = sorted((sumn,sump))
    if abs(sum) <= Tolerance*min : return (type(Tolerance))(0)
    return sum

Major axis

[edit | edit source]
# axis is perpendicular to directrix.
ax,bx = b1,-a1
# axis contains foci. ax + by + c = 0
cx = reduce_Decimal_number(-(ax*p1 + bx*q1))
axis = ax,bx,cx
print ( '    Axis : ({})x + ({})y + ({}) = 0'.format(ax,bx,cx) )
print ( '    Eccentricity = {}'.format(e1) )
print ()
print ( '    Directrix 1 : ({})x + ({})y + ({}) = 0'.format(a1,b1,c1) )
print ( '    Directrix 2 : ({})x + ({})y + ({}) = 0'.format(a2,b2,c2) )
print ( '    Distance between directrices = {}'.format(abs(c1-c2)) )
F1 = p1,q1 # Focus 1.
print ( '    F1 : ({}, {})'.format(p1,q1) )
F2 = p2,q2 # Focus 2.
print ( '    F2 : ({}, {})'.format(p2,q2) )

#  Direction cosines along axis from F1 towards F2:
dx,dy = a1,b1

# p2 = p1 + dx*distance_F1_F2
# q2 = q1 + dy*distance_F1_F2
if dx : distance_F1_F2 = (p2 - p1)/dx
else :  distance_F1_F2 = (q2 - q1)
if distance_F1_F2 < 0 :
    distance_F1_F2 *= -1
    dx *= -1 ; dy *= -1
print ( '    Distance between foci = {}'.format(distance_F1_F2) )

# Intercept on directrix1
distance_from_F1_to_ID1 = abs(a1*p1 + b1*q1 + c1)
ID1 = xID1,yID1 = p1 - dx*distance_from_F1_to_ID1, q1 - dy*distance_from_F1_to_ID1
print ( '    Intercept ID1 : ({}, {})'.format(xID1,yID1) )

# 
# distance_F1_F2
# -------------------- = e
# length_of_major_axis
# 
length_of_major_axis = distance_F1_F2 / e1

# Intercept1 on curve
distance_from_F1_to_curve = (length_of_major_axis - distance_F1_F2 )/2

xI1,yI1 = p1 - dx*distance_from_F1_to_curve, q1 - dy*distance_from_F1_to_curve
I1 = xI1,yI1 = [ reduce_Decimal_number(v) for v in (xI1,yI1) ]
print ( '    Intercept I1 : ({}, {})'.format(xI1,yI1) )
    Axis : (-0.8)x + (-0.6)y + (9.4) = 0
    Eccentricity = 0.9

    Directrix 1 : (0.6)x + (-0.8)y + (-27.47368421052631578947) = 0
    Directrix 2 : (0.6)x + (-0.8)y + (2) = 0
    Distance between directrices = 29.47368421052631578947
    F1 : (22.32421052631578947368, -14.09894736842105263158)
    F2 : (8, 5)
    Distance between foci = 23.87368421052631578947
    Intercept ID1 : (24.00421052631578947368, -16.33894736842105263158)
    Intercept I1 : (23.12, -15.16)

Techniques similar to above can be used to calculate points

Latus rectums

[edit | edit source]
# direction cosines along latus rectum.
dlx,dly = -dy,dx
# 
# distance from U to F1               half_latus_rectum
# ------------------------------ = ----------------------- = e1
# distance from U to directrix 1   distance_from_F1_to_ID1
# 
half_latus_rectum = reduce_Decimal_number(e1*distance_from_F1_to_ID1)
# latus rectum 1
# Focal chord has equation (afc)x + (bfc)y + (cfc) = 0.
afc,bfc = a1,b1
cfc = reduce_Decimal_number(-(afc*p1 + bfc*q1))
print ( '    Focal chord PU : ({})x + ({})y + ({}) = 0'.format(afc,bfc,cfc) )
        
P = xP,yP = p1 + dlx*half_latus_rectum, q1 + dly*half_latus_rectum
print ( '    Point P : ({}, {})'.format(xP,yP) )
U = xU,yU = p1 - dlx*half_latus_rectum, q1 - dly*half_latus_rectum
print ( '    Point U : ({}, {})'.format(xU,yU) )
distance = reduce_Decimal_number(( (xP - xU)**2 + (yP - yU)**2 ).sqrt())
print ('    Length PU =', distance)
print ('    half_latus_rectum =', half_latus_rectum)
    Focal chord PU : (0.6)x + (-0.8)y + (-24.67368421052631578947) = 0
    Point P : (20.30821052631578947368, -15.61094736842105263158)
    Point U : (24.34021052631578947368, -12.58694736842105263158)
    Length PU = 5.04
    half_latus_rectum = 2.52

Techniques similar to above can be used to calculate points

Minor axis

[edit | edit source]
print ()

# Mid point between F1, F2:
M = xM,yM = (p1 + p2)/2, (q1 + q2)/2
print ( '    Mid point M : ({}, {})'.format(xM,yM) )
half_major  = length_of_major_axis / 2
half_distance = distance_F1_F2 / 2

# half_distance**2 + half_minor**2 = half_major**2
half_minor = ( half_major**2 - half_distance**2 ).sqrt()
length_of_minor_axis = half_minor * 2
        
Q = xQ,yQ = xM + dlx*half_minor,  yM + dly*half_minor
T = xT,yT = xM - dlx*half_minor,  yM - dly*half_minor
print ( '    Point Q : ({}, {})'.format(xQ,yQ) )
print ( '    Point T : ({}, {})'.format(xT,yT) )

print ('    length_of_major_axis =', length_of_major_axis)
print ('    length_of_minor_axis =', length_of_minor_axis)
# 
# A basic check.
# length_of_minor_axis**2 = (length_of_major_axis**2)(1-e**2)
# 
# length_of_minor_axis**2
# ----------------------- = 1-e**2
# length_of_major_axis**2
# 
# length_of_minor_axis**2
# ----------------------- + (e**2 - 1) = 0
# length_of_major_axis**2
# 
values = (length_of_minor_axis/length_of_major_axis)**2, e1**2 - 1
sum_zero(values) and 3/0
        
aM,bM = a1,b1 # Minor axis is parallel to directrix.
cM = reduce_Decimal_number(-(aM*xM + bM*yM))
print ( '    Minor axis : ({})x + ({})y + ({}) = 0'.format(aM,bM,cM) )
    Mid point M : (15.16210526315789473684, -4.54947368421052631579)
    Point Q : (10.53708406832736953616, -8.018239580333420216299)
    Point T : (19.78712645798841993752, -1.080707788087632415281)
    length_of_major_axis = 26.52631578947368421052
    length_of_minor_axis = 11.56255298707631300170
    Minor axis : (0.6)x + (-0.8)y + (-12.73684210526315789474) = 0

Checking

[edit | edit source]

All interesting points have been calculated without using equations of any of the relevant lines.

However, equations of relevant lines are very useful for testing, for example:

  • Check that points are on axis.
  • Check that points are on latus rectum through
  • Check that points are on minor axis through
  • Check that points are on latus rectum through


Test below checks that 8 points are on ellipse and satisfy eccentricity

t1 = (
    ('I1'), ('I2'),
    ('P'), ('Q'), ('R'),
    ('S'), ('T'), ('U'),
)

for name in t1 :
    value = eval(name)
    x,y = [ reduce_Decimal_number(v) for v in value ]
    print ('{} : ({}, {})'.format((name+' ')[:2], x,y))
    values = A*x**2, B*y**2, C*x*y, D*x, E*y, F
    sum_zero(values) and 3/0

    # Relative to Directrix 1 and Focus 1:
    distance_to_F1 = ( (x-p1)**2 + (y-q1)**2 ).sqrt()
    distance_to_directrix1 = a1*x + b1*y + c1
    e1 = distance_to_F1 / distance_to_directrix1
    print ('    e1 =',e1) # Raw value is printed.

    # Relative to Directrix 2 and Focus 2:
    distance_to_F2 = ( (x-p2)**2 + (y-q2)**2 ).sqrt()
    distance_to_directrix2 = a2*x + b2*y + c2
    e2 = distance_to_F2 / distance_to_directrix2
    e2 = reduce_Decimal_number(e2)
    print ('    e2 =',e2) # Clean value is printed.

Note the differences between "raw" values of and "clean" values of

I1 : (23.12, -15.16)
    e1 = -0.9000000000000000000034 
    e2 = 0.9
I2 : (7.204210526315789473684, 6.061052631578947368421)
    e1 = -0.9 
    e2 = 0.9
P  : (20.30821052631578947368, -15.61094736842105263158) 
    e1 = -0.9 
    e2 = 0.9
Q  : (10.53708406832736953616, -8.018239580333420216299)
    e1 = -0.9000000000000000000002
    e2 = 0.9
R  : (5.984, 3.488)
    e1 = -0.9000000000000000000003
    e2 = 0.9
S  : (10.016, 6.512) 
    e1 = -0.9000000000000000000003
    e2 = 0.9   
T  : (19.78712645798841993752, -1.080707788087632415281)
    e1 = -0.8999999999999999999996
    e2 = 0.9
U  : (24.34021052631578947368, -12.58694736842105263158)
    e1 = -0.9
    e2 = 0.9

Other resources

[edit | edit source]
  • Should the contents of this Wikiversity page be merged into the related Wikibooks modules such as b:Conic Sections/Ellipse?