# Conic sections

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

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

${\displaystyle {\frac {(x-h)^{2}}{a^{2}}}+{\frac {(y-k)^{2}}{b^{2}}}=1}$    (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(${\displaystyle x_{1}}$, ${\displaystyle y_{1}}$) tangents will have the equation:

${\displaystyle {\frac {xx_{1}}{a^{2}}}+{\frac {yy_{1}}{b^{2}}}=1}$

And normals:

${\displaystyle {\frac {xa^{2}}{x_{1}}}-{\frac {yb^{2}}{y_{1}}}=a^{2}-b^{2}}$

Likewise for the parametric coordinates of P, (a ${\displaystyle \cos \alpha }$, b ${\displaystyle \sin \alpha }$),

${\displaystyle {\frac {x\cos \alpha }{a}}+{\frac {y\sin \alpha }{b}}=1}$

## Properties of Ellipses

S and S' are typically regarded as the two foci of the ellipse. Where ${\displaystyle a>b}$, these become (ae, 0) and (-ae, 0) respectively. Where ${\displaystyle a these become (0, be) and (0, -be) respectively.

A point P on the ellipse will move about these two foci ut ${\displaystyle |PS+PS'|=2a}$

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

${\displaystyle b^{2}=a^{2}(1-e^{2})}$

The directrix will be: ${\displaystyle x=\pm {\frac {a}{e}}}$

## Circles

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

Hence, the equation becomes:

${\displaystyle x^{2}+y^{2}=r^{2}}$

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

## Hyperbolas

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

Centered at the origin, Hyperbolas have the general equation:

${\displaystyle {\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1}$

A point P on will move about the two foci ut ${\displaystyle |PS-PS'|=2a}$

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

From a point P(${\displaystyle x_{1}}$, ${\displaystyle y_{1}}$) tangents will have the equation:

${\displaystyle {\frac {xx_{1}}{a^{2}}}-{\frac {yy_{1}}{b^{2}}}=1}$

And normals:

${\displaystyle {\frac {xa^{2}}{x_{1}}}+{\frac {yb^{2}}{y_{1}}}=a^{2}+b^{2}}$

The directrixes (singular directrix) and foci of hyperbolas are the same as those of ellipses, namely directrixes of ${\displaystyle x=\pm {\frac {a}{e}}}$ and foci of ${\displaystyle (\pm ae,0)}$

The asymptotes of a hyperbola lie at ${\displaystyle y=\pm {\frac {b}{a}}x}$

## Rectangular Hyperbolas

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

${\displaystyle xy=c}$

## Conic sections generally

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

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

Let the point have coordinates ${\displaystyle (x,y).}$

Let the focus have coordinates ${\displaystyle (p,q).}$

Let the directrix have equation ${\displaystyle ax+by+c=0}$ where ${\displaystyle a^{2}+b^{2}=1.}$

Then ${\displaystyle e={\frac {\text{distance to focus}}{\text{distance to directrix}}}}$ ${\displaystyle ={\frac {\sqrt {(x-p)^{2}+(y-q)^{2}}}{ax+by+c}}}$

${\displaystyle e(ax+by+c)={\sqrt {(x-p)^{2}+(y-q)^{2}}}}$

Square both sides: ${\displaystyle (ax+by+c)(ax+by+c)e^{2}=(x-p)^{2}+(y-q)^{2}}$

Rearrange: ${\displaystyle (x-p)^{2}+(y-q)^{2}-(ax+by+c)(ax+by+c)e^{2}=0\ \dots \ (1).}$

Expand ${\displaystyle (1),}$ simplify, gather like terms and result is:

${\displaystyle Ax^{2}+By^{2}+Cxy+Dx+Ey+F=0}$ where:

${\displaystyle X=e^{2}}$

${\displaystyle A=Xa^{2}-1}$

${\displaystyle B=Xb^{2}-1}$

${\displaystyle C=2Xab}$

${\displaystyle D=2p+2Xac}$

${\displaystyle E=2q+2Xbc}$

${\displaystyle F=Xc^{2}-p^{2}-q^{2}}$

 Note that values ${\displaystyle A,B,C,D,E,F}$ depend on: ${\displaystyle e}$ non-zero. This method is not suitable for circle where ${\displaystyle e=0.}$ ${\displaystyle e^{2}.}$ Sign of ${\displaystyle e\pm }$ is not significant. ${\displaystyle (ax+by+c)^{2}.\ ((-a)x+(-b)y+(-c))^{2}}$ or ${\displaystyle ((-1)(ax+by+c))^{2}}$ and ${\displaystyle (ax+by+c)^{2}}$ produce same result. For example, directrix ${\displaystyle 0.6x-0.8y+3=0}$ and directrix ${\displaystyle -0.6x+0.8y-3=0}$ produce same result.

### Implementation

# 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

#### Parabola

Every parabola has eccentricity ${\displaystyle e=1.}$

 Quadratic function complies with definition of parabola. Distance from point ${\displaystyle (6,9)}$ to focus = distance from point ${\displaystyle (6,9)}$ to directrix = 10. Distance from point ${\displaystyle (0,0)}$ to focus = distance from point ${\displaystyle (0,0)}$ to directrix = 1. Simple quadratic function: Let focus be point ${\displaystyle (0,1).}$ Let directrix have equation: ${\displaystyle y=-1}$ or ${\displaystyle (0)x+(1)y+1=0.}$ # 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: ${\displaystyle (-1)x^{2}+(0)y^{2}+(0)xy+(0)x+(4)y+(0)=0}$ Curve is quadratic function: ${\displaystyle 4y=x^{2}}$ or ${\displaystyle y={\frac {x^{2}}{4}}}$ 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
 Curve in Figure 1 below has: Directrix: ${\displaystyle y=-23}$ Focus: ${\displaystyle (7,-21)}$ Equation: ${\displaystyle (-1)x^{2}+(0)y^{2}+(0)xy+(14)x+(4)y+(39)=0}$ or ${\displaystyle y={\frac {x^{2}-14x-39}{4}}}$ Curve in Figure 2 below has: Directrix: ${\displaystyle x=12}$ Focus: ${\displaystyle (10,-7)}$ Equation: ${\displaystyle (0)x^{2}+(-1)y^{2}+(0)xy+(-4)x+(-14)y+(-5)=0}$ or ${\displaystyle x={\frac {-(y^{2}+14y+5)}{4}}}$ Curve in Figure 3 below has: Directrix: ${\displaystyle (0.6)x-(0.8)y+(2.0)=0}$ Focus: ${\displaystyle (6.6,6.2)}$ Equation: ${\displaystyle -(0.64)x^{2}-(0.36)y^{2}-(0.96)xy+(15.6)x+(9.2)y-(78)=0}$ Figure 1.${\displaystyle y={\frac {x^{2}-14x-39}{4}}}$ Figure 2.${\displaystyle x={\frac {-(y^{2}+14y+5)}{4}}}$ Figure 3.${\displaystyle -(0.64)x^{2}-(0.36)y^{2}}$${\displaystyle -(0.96)xy+(15.6)x}$${\displaystyle +(9.2)y-(78)=0}$

#### Ellipse

Every ellipse has eccentricity ${\displaystyle 1>e>0.}$

 Ellipse with ecccentricity of 0.25 and center at origin. Point1 ${\displaystyle =(0,3.87298334620741688517926539978).}$ Eccentricity ${\displaystyle e={\frac {\text{distance from point1 to focus}}{\text{distance from point1 to directrix}}}={\frac {4}{16}}=0.25.}$ For every point on curve, ${\displaystyle e=0.25.}$ A simple ellipse: Let focus be point ${\displaystyle (p,q)}$ where ${\displaystyle p,q=-1,0}$ Let directrix have equation: ${\displaystyle (1)x+(0)y+16=0}$ or ${\displaystyle x=-16.}$ Let eccentricity ${\displaystyle e=0.25}$ # 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: ${\displaystyle (0.9375)x^{2}+(1)y^{2}=(15).}$ 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 ${\displaystyle 0,}$ shape of ellipse approaches shape of circle. As eccentricity approaches ${\displaystyle 1,}$ shape of ellipse approaches shape of parabola. The effect of eccentricity. All ellipses in diagram have: Focus at point ${\displaystyle (-1,0)}$ Directrix with equation ${\displaystyle x=-16.}$ Five ellipses are shown with eccentricities varying from ${\displaystyle 0.1}$ to ${\displaystyle 0.9.}$
 Curve in Figure 1 below has: Directrix: ${\displaystyle x=-10}$ Focus: ${\displaystyle (3,0)}$ Eccentricity: ${\displaystyle e=0.5}$ Equation: ${\displaystyle (-0.75)x^{2}+(-1)y^{2}+(0)xy+(11)x+(0)y+(16)=0}$ Curve in Figure 2 below has: Directrix: ${\displaystyle y=-12}$ Focus: ${\displaystyle (7,-4)}$ Eccentricity: ${\displaystyle e=0.7}$ Equation: ${\displaystyle (-1)x^{2}+(-0.51)y^{2}+(0)xy+(14)x+(3.76)y+(5.56)=0}$ Curve in Figure 3 below has: Directrix: ${\displaystyle (0.6)x-(0.8)y+(2.0)=0}$ Focus: ${\displaystyle (8,5)}$ Eccentricity: ${\displaystyle e=0.9}$ Equation: ${\displaystyle (-0.7084)x^{2}+(-0.4816)y^{2}+(-0.7776)xy+(17.944)x+(7.408)y+(-85.76)=0}$ Figure 1.Ellipse on X axis. Figure 2.Ellipse parallel to Y axis. Figure 3.Ellipse with random orientation.

#### Hyperbola

Every hyperbola has eccentricity ${\displaystyle e>1.}$

 Hyperbola with eccentricity of 1.5 and center at origin. Point1 ${\displaystyle =(22.5,21).}$ Eccentricity ${\displaystyle e={\frac {\text{distance from point1 to focus}}{\text{distance from point1 to directrix}}}={\frac {37.5}{25}}=1.5.}$ For every point on curve, ${\displaystyle e=1.5.}$ A simple hyperbola: Let focus be point ${\displaystyle (p,q)}$ where ${\displaystyle p,q=0,-9}$ Let directrix have equation: ${\displaystyle (0)x+(1)y+4=0}$ or ${\displaystyle y=-4.}$ Let eccentricity ${\displaystyle e=1.5}$ # 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: ${\displaystyle (1.25)y^{2}-x^{2}=45.}$ 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: ${\displaystyle y=-4.}$ As eccentricity approaches ${\displaystyle 1,}$ shape of curve approaches shape of parabola. The effect of eccentricity. All hyperbolas in diagram have: Focus at point ${\displaystyle (0,-9)}$ Directrix with equation ${\displaystyle y=-4.}$ Six hyperbolas are shown with eccentricities varying from ${\displaystyle 1.5}$ to ${\displaystyle 20.}$
 Curve in Figure 1 below has: Directrix: ${\displaystyle y=6}$ Focus: ${\displaystyle (0,1)}$ Eccentricity: ${\displaystyle e=1.5}$ Equation: ${\displaystyle (-1)x^{2}+(1.25)y^{2}+(0)xy+(0)x+(-25)y+(80)=0}$ Curve in Figure 2 below has: Directrix: ${\displaystyle x=1}$ Focus: ${\displaystyle (-5,6)}$ Eccentricity: ${\displaystyle e=2.5}$ Equation: ${\displaystyle (5.25)x^{2}+(-1)y^{2}+(0)xy+(-22.5)x+(12)y+(-54.75)=0}$ Curve in Figure 3 below has: Directrix: ${\displaystyle (0.8)x+(0.6)y+(2.0)=0}$ Focus: ${\displaystyle (-28,12)}$ Eccentricity: ${\displaystyle e=1.2}$ Equation: ${\displaystyle (-0.0784)x^{2}+(-0.4816)y^{2}+(1.3824)xy+(-51.392)x+(27.456)y+(-922.24)=0}$ Figure 1.Hyperbola on Y axis. Figure 2.Hyperbola parallel to X axis. Figure 3.Hyperbola with random orientation.

### Reversing the process

The expression "reversing the process" means calculating the values of ${\displaystyle e,}$ focus and directrix when given the equation of the conic section, the familiar values ${\displaystyle A,B,C,D,E,F.}$

Consider the equation of a simple ellipse: ${\displaystyle 0.9375x^{2}+y^{2}=15.}$ This is a conic section where ${\displaystyle A,B,C,D,E,F=-0.9375,-1,0,0,0,15.}$

This ellipse may be expressed as ${\displaystyle 15x^{2}+16y^{2}=240,}$ a format more appealing to the eye than numbers containing fractions or decimals.

However, when this ellipse is expressed as ${\displaystyle -0.9375x^{2}-y^{2}+15=0,}$ this format is the ellipse expressed in "standard form," a notation that greatly simplifies the calculation of ${\displaystyle a,b,c,e,p,q.}$

 Modify the equations for ${\displaystyle A,B,C}$ slightly: ${\displaystyle KA=Xaa-1}$ or ${\displaystyle Xaa=KA+1\ \dots \ (1)}$ ${\displaystyle KB=Xbb-1}$ or ${\displaystyle Xbb=KB+1\ \dots \ (2)}$ ${\displaystyle KC=2Xab\ \dots \ (3)}$ ${\displaystyle (3)\ {\text{squared:}}\ KKCC=4XaaXbb\ \dots \ (4)}$ In ${\displaystyle (4)}$ substitute for ${\displaystyle Xaa,Xbb:}$ ${\displaystyle C^{2}K^{2}=4(KA+1)(KB+1)\ \dots \ (5)}$ ${\displaystyle (5)}$ is a quadratic equation in ${\displaystyle K:\ (a\_)K^{2}+(b\_)K+(c\_)=0}$ where: ${\displaystyle a\_=4AB-C^{2}}$ ${\displaystyle b\_=4(A+B)}$ ${\displaystyle c\_=4}$ Because ${\displaystyle (5)}$ is a quadratic equation, the solution of ${\displaystyle (5)}$ may contain an unwanted value of ${\displaystyle K}$ that will be eliminated later. From ${\displaystyle (1)}$ and ${\displaystyle (2):}$ ${\displaystyle Xaa+Xbb=KA+KB+2}$ ${\displaystyle X(aa+bb)=KA+KB+2}$ Because ${\displaystyle aa+bb=1,\ X=KA+KB+2}$

#### Implementation

 # 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

 The values ${\displaystyle D,E,F:}$ ${\displaystyle D=2p+2Xac;\ 2p=(D-2Xac)}$ ${\displaystyle E=2q+2Xbc;\ 2q=(E-2Xbc)}$ ${\displaystyle F=Xcc-pp-qq\ \dots \ (10)}$ ${\displaystyle (10)*4:\ 4F=4Xcc-4pp-4qq\ \dots \ (11)}$ In ${\displaystyle (11)}$ replace ${\displaystyle 4pp,4qq:\ 4F=4Xcc-(D-2Xac)(D-2Xac)-(E-2Xbc)(E-2Xbc)\ \dots \ (12)}$ Expand ${\displaystyle (12),}$ simplify, gather like terms and result is quadratic function in ${\displaystyle c:}$ ${\displaystyle (a\_)c^{2}+(b\_)c+(c\_)=0\ \dots \ (14)}$ where: ${\displaystyle a\_=4X(1-Xaa-Xbb)}$ ${\displaystyle aa+bb=1,}$ Therefore: ${\displaystyle a\_=4X(1-X)}$ ${\displaystyle b\_=4X(Da+Eb)}$ ${\displaystyle c\_=-(D^{2}+E^{2}+4F)}$ For parabola, there is one value of ${\displaystyle c}$ because there is one directrix. For ellipse and hyperbola, there are two values of ${\displaystyle c}$ because there are two directrices.

#### Implementation

 # 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

#### Parabola

 Graph of parabola ${\displaystyle 16x^{2}+9y^{2}-24xy+410x-420y+3175=0.}$ Equation of parabola is given. This section calculates ${\displaystyle {\text{eccentricity, focus, directrix.}}}$ Given equation of conic section: ${\displaystyle 16x^{2}+9y^{2}-24xy+410x-420y+3175=0.}$ Calculate ${\displaystyle {\text{eccentricity, focus, directrix.}}}$ # 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: ${\displaystyle 0.6x+0.8y+3=0}$ Eccentricity: ${\displaystyle e=1}$ Focus: ${\displaystyle p,q=-10,6}$ Because eccentricity is ${\displaystyle 1,}$ 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

 Graph of ellipse ${\displaystyle 481x^{2}+369y^{2}-384xy+5190x+5670y+7650=0.}$ Equation of ellipse is given. This section calculates ${\displaystyle {\text{eccentricity, foci, directrices.}}}$ Given equation of conic section: ${\displaystyle 481x^{2}+369y^{2}-384xy+5190x+5670y+7650=0.}$ Calculate ${\displaystyle {\text{eccentricity, foci, directrices.}}}$ # 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: ${\displaystyle 0.6x+0.8y-3=0}$ Eccentricity: ${\displaystyle e=0.8}$ Focus 1: ${\displaystyle p,q=-3,-3}$ Directrix 2: ${\displaystyle 0.6x+0.8y+37=0}$ Eccentricity: ${\displaystyle e=0.8}$ Focus 2: ${\displaystyle p,q=-18.36,-23.48}$ Because eccentricity is ${\displaystyle 0.8,}$ 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

 Graph of hyperbola ${\displaystyle 7x^{2}+0y^{2}-24xy+90x+216y-81=0.}$ Equation of hyperbola is given. This section calculates ${\displaystyle {\text{eccentricity, foci, directrices.}}}$ Given equation of conic section: ${\displaystyle 7x^{2}+0y^{2}-24xy+90x+216y-81=0.}$ Calculate ${\displaystyle {\text{eccentricity, foci, directrices.}}}$ # 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: ${\displaystyle 0.6x+0.8y-3=0}$ Eccentricity: ${\displaystyle e=1.25}$ Focus 1: ${\displaystyle p,q=0,-3}$ Directrix 2: ${\displaystyle 0.6x+0.8y-22.2=0}$ Eccentricity: ${\displaystyle e=1.25}$ Focus 2: ${\displaystyle p,q=18,21}$ Because eccentricity is ${\displaystyle 1.25,}$ 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

Given equation of conic section: ${\displaystyle Ax^{2}+By^{2}+Cxy+Dx+Ey+F=0,}$

differentiate both sides with respect to ${\displaystyle x.}$

${\displaystyle 2Ax+B(2yy')+C(xy'+y)+D+Ey'=0}$

${\displaystyle 2Ax+2Byy'+Cxy'+Cy+D+Ey'=0}$

${\displaystyle 2Byy'+Cxy'+Ey'+2Ax+Cy+D=0}$

${\displaystyle y'(2By+Cx+E)=-(2Ax+Cy+D)}$

${\displaystyle y'={\frac {-(2Ax+Cy+D)}{Cx+2By+E}}}$

For slope horizontal: ${\displaystyle 2Ax+Cy+D=0.}$

For slope vertical: ${\displaystyle Cx+2By+E=0.}$

For given slope ${\displaystyle m={\frac {-(2Ax+Cy+D)}{Cx+2By+E}}}$

${\displaystyle m(Cx+2By+E)=-2Ax-Cy-D}$

${\displaystyle mCx+2Ax+m2By+Cy+mE+D=0}$

${\displaystyle (mC+2A)x+(m2B+C)y+(mE+D)=0.}$

### Implementation

 # 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

##### y = f(x)
 Graph of quadratic function ${\displaystyle y={\frac {x^{2}-14x-39}{4}}.}$ At intersection of ${\displaystyle {\text{line 1}}}$ and curve, slope = ${\displaystyle 0}$. At intersection of ${\displaystyle {\text{line 2}}}$ and curve, slope = ${\displaystyle 5}$. Slope of curve is never vertical. Consider conic section: ${\displaystyle (-1)x^{2}+(0)y^{2}+(0)xy+(14)x+(4)y+(39)=0.}$ This is quadratic function: ${\displaystyle y={\frac {x^{2}-14x-39}{4}}}$ Slope of this curve: ${\displaystyle m=y'={\frac {2x-14}{4}}}$ Produce values for slope horizontal, slope vertical and slope ${\displaystyle 5:}$ ${\displaystyle }$${\displaystyle }$${\displaystyle }$${\displaystyle }$${\displaystyle }$ # 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)
 Graph of quadratic function ${\displaystyle x={\frac {-(y^{2}+14y+5)}{4}}.}$ At intersection of ${\displaystyle {\text{line 1}}}$ and curve, slope is vertical. At intersection of ${\displaystyle {\text{line 2}}}$ and curve, slope = ${\displaystyle 0.5}$. Slope of curve is never horizontal. Consider conic section: ${\displaystyle (0)x^{2}+(-1)y^{2}+(0)xy+(-4)x+(-14)y+(-5)=0.}$ This is quadratic function: ${\displaystyle x={\frac {-(y^{2}+14y+5)}{4}}}$ Slope of this curve: ${\displaystyle {\frac {dx}{dy}}={\frac {-2y-14}{4}}}$ ${\displaystyle m=y'={\frac {dy}{dx}}={\frac {-4}{2y+14}}}$ Produce values for slope horizontal, slope vertical and slope ${\displaystyle 0.5:}$ ${\displaystyle }$${\displaystyle }$${\displaystyle }$${\displaystyle }$${\displaystyle }$ # 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

 Graph of parabola ${\displaystyle (9)x^{2}+(16)y^{2}+(-24)xy+(104)x+(28)y+(-144)=0.}$ At intersection of ${\displaystyle {\text{Line 1}}}$ and curve, slope is horizontal. At intersection of ${\displaystyle {\text{Line 2}}}$ and curve, slope is vertical. At intersection of ${\displaystyle {\text{Line 3}}}$ and curve, slope = ${\displaystyle 2}$. Slope of curve is never ${\displaystyle 0.75}$ because axis has slope ${\displaystyle 0.75}$ and curve is never parallel to axis. Consider conic section: ${\displaystyle (9)x^{2}+(16)y^{2}+(-24)xy+(104)x+(28)y+(-144)=0.}$ This curve is a parabola. Produce values for slope horizontal, slope vertical and slope ${\displaystyle 2:}$ # 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 ${\displaystyle {\frac {3}{4}}.}$ Produce values for slope horizontal, slope vertical and slope ${\displaystyle 0.75:}$ # 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 ${\displaystyle 0.75}$ and curve is never parallel to axis.

#### Ellipse

 Graph of ellipse ${\displaystyle (1771)x^{2}+(1204)y^{2}+(1944)xy+(-44860)x+(-18520)y+(214400)=0.}$ At intersection of ${\displaystyle {\text{Line 1}}}$ and curve, slope is horizontal. At intersection of ${\displaystyle {\text{Line 2}}}$ and curve, slope is vertical. At intersection of ${\displaystyle {\text{Line 3}}}$ and curve, slope = ${\displaystyle -1.}$ Consider conic section: ${\displaystyle (1771)x^{2}+(1204)y^{2}+(1944)xy+(-44860)x+(-18520)y+(214400)=0.}$ This curve is an ellipse. Produce values for slope horizontal, slope vertical and slope ${\displaystyle -1:}$ # 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 ${\displaystyle {\frac {1}{0}}.}$ If slope of curve is given as ${\displaystyle {\frac {1}{0}},}$ it means that curve is vertical at that point and tangent to curve has equation ${\displaystyle x=k.}$ 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

 Graph of hyperbola ${\displaystyle (-351)x^{2}+(176)y^{2}+(-336)xy+(4182)x+(-3824)y+(-16231)=0.}$ At intersection of ${\displaystyle {\text{Line 1}}}$ and curve, slope is horizontal. ${\displaystyle {\text{Line 2}}}$ and curve do not intersect. Slope is never vertical. At intersection of ${\displaystyle {\text{Line 3}}}$ and curve, slope = ${\displaystyle 2.}$ Consider conic section: ${\displaystyle (-351)x^{2}+(176)y^{2}+(-336)xy+(4182)x+(-3824)y+(-16231)=0.}$ This curve is a hyperbola. Produce values for slope horizontal, slope vertical and slope ${\displaystyle 2:}$ # 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

## Asymptotes of Hyperbola

Let the conic section have the familiar equation: ${\displaystyle Ax^{2}+By^{2}+Cxy+Dx+Ey+F=0.}$

Let a line have equation: ${\displaystyle y=f(x)=mx+c.}$

Let ${\displaystyle f(x)}$ intersect the conic section. For ${\displaystyle y}$ in equation of conic section substitute ${\displaystyle (mx+c).}$

${\displaystyle y=g(x)=Ax^{2}+B(mx+c)^{2}+Cx(mx+c)+Dx+E(mx+c)+F=0.}$

Expand ${\displaystyle g(x),}$ simplify, gather like terms and result is quadratic function in ${\displaystyle x:}$

${\displaystyle h(x)=(a}$_${\displaystyle )x^{2}+(b}$_${\displaystyle )x+(c}$_${\displaystyle )=0}$ where:

${\displaystyle a}$_ ${\displaystyle =A+Bm^{2}+Cm}$

${\displaystyle b}$_ ${\displaystyle =2Bcm+cC+D+Em}$

${\displaystyle c}$_ ${\displaystyle =Ec+F+Bc^{2}}$

If line ${\displaystyle f(x)}$ is an asymptote, then ${\displaystyle a}$_ ${\displaystyle =b}$_ ${\displaystyle =0,}$ in which case:

${\displaystyle j(m)=\ (}$_${\displaystyle a)m^{2}+(}$_${\displaystyle b)m+(}$_${\displaystyle c)=0}$ where:

_${\displaystyle a=B;\ }$_${\displaystyle b=C;\ }$_${\displaystyle c=A}$

and ${\displaystyle m_{1},m_{2}}$ (roots of ${\displaystyle j(m)}$) are the slopes of the 2 asymptotes.

### Implementation

# python code

def calculate_asymptotes (ABCDEF, XY, flag = 0) :
'''
line1,line2 = calculate_asymptotes (ABCDEF, XY[, flag])
where each line is (a,b,c) meaning ax + by + c = 0.
XY is mid-point of line joining 2 foci.
'''
temp = []
A,B,C,D,E,F = [ dD(str(v)) for v in ABCDEF ]
_a,_b,_c = B,C,A
X,Y = XY

if _a == 0 :
# Allow for the possibility that one asymptote may be a vertical line.
m = 1,0  # m = 1/0
temp += [m]
# _b m + _c = 0
#
#     -_c
# m = ---
#      _b
#
m = -_c,_b
temp += [m]
else :
discr = _b**2 - 4*_a*_c
root = discr.sqrt()
top1,top2 = (-_b - root),  (-_b + root)
bottom = 2*_a
m = top1, bottom
temp += [m]
m = top2, bottom
temp += [m]

output = []
for p,q in temp :
#     p
# y = - x + c
#     q
#
# qy = px + qc
# px - qy + qc = 0
# a1 x + b1 y + c1 = 0
a1,b1 = p,-q
# line contains mid_point. Using c = - (ax+by)
c1 = -(a1*X + b1*Y)
abc = tuple( reduce_Decimal_number(v) for v in (a1,b1,c1) )
output += [abc]

if flag :
# Print output and check results.
for a1,b1,c1 in output :
print ('asymptote: ({})x + ({})y + ({}) = 0'.format(a1,b1,c1))
if not b1 : continue
def almost_zero(input) :
#
# This function adds all values in input and
# verifies that sum is acceptably close to 0
# given the sizes of values in input.
#
sum = sump = 0
for v in input :
sum += v
if v > 0 : sump += v
# sump + sumn = sum
sumn = abs(sum-sump)
min,max = sorted((sump,sumn))
return (abs(sum) <= Tolerance*min)
#
# a1 x + b1 y + c1 = 0
# b1 y = -a1 x - c1
#     -a1     -c1
# y = --- x + ---
#      b1      b1
#
m = -a1/b1 ; c = -c1/b1
# a_ = A + Bmm + Cm. a_ should be 0.
almost_zero((A, B*m**2, C*m)) or 1/0
# b_ = 2Bcm + cC + D + Em. b_ should be 0.
almost_zero((2*B*c*m, c*C, D, E*m)) or 2/0

return output
 Figure 1.Hyperbola parallel to X axis. Figure 2.Hyperbola with asymptote horizontal. Figure 3.Hyperbola with asymptote vertical. Figure 4.Hyperbola with random orientation.

## Latera recta et cetera

"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 ${\displaystyle 1771x^{2}+1204y^{2}+1944xy-44860x-18520y+214400=0.}$ 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 ${\displaystyle {\text{ID2}}}$ = (6.32, 7.24) ${\displaystyle {\text{I2}}}$ = (7.204210526315789473684, 6.061052631578947368421) F2 = (8, 5) M = (15.16210526315789473684, -4.54947368421052631579) F1 = (22.32421052631578947368, -14.09894736842105263158) ${\displaystyle {\text{I1}}}$ = (23.12, -15.16) ${\displaystyle {\text{ID1}}}$ = (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: ${\displaystyle {\text{ID1ID2}}}$ = 29.47368421052631578947 Length of major axis: ${\displaystyle {\text{I1I2}}}$ = 26.52631578947368421052 Distance between foci: ${\displaystyle {\text{F1F2}}}$ = 23.87368421052631578947 Length of minor axis: QT = 11.56255298707631300170 Length of latus rectum: RS = PU = 5.04 Consider conic section: ${\displaystyle 1771x^{2}+1204y^{2}+1944xy-44860x-18520y+214400=0.}$ 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

# 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 ${\displaystyle I2,ID2.}$

### Latus rectums

# 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 ${\displaystyle R,S.}$

### Minor axis

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

 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 ${\displaystyle ID2,I2,F2,M,F1,I1,ID1}$ are on axis. Check that points ${\displaystyle R,F2,S}$ are on latus rectum through ${\displaystyle F2.}$ Check that points ${\displaystyle Q,M,T}$ are on minor axis through ${\displaystyle M.}$ Check that points ${\displaystyle P,F1,U}$ are on latus rectum through ${\displaystyle F1.}$ Test below checks that 8 points ${\displaystyle I1,I2,P,Q,R,S,T,U}$ are on ellipse and satisfy eccentricity ${\displaystyle e=0.9.}$
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 ${\displaystyle e_{1}}$ and "clean" values of ${\displaystyle e_{2}.}$
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

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