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.
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.
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:
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:
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.
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.
# python codeimportdecimaldD=decimal.Decimal# Decimal object is like a float with (almost) unlimited precision.dgt=decimal.getcontext()Precision=dgt.prec=22defreduce_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) :'iftype(number)!=dD:number=dD(str(number))f1=float(number)if(f1+1)==1:returndD(0)ifint(f1)==f1:returndD(int(f1))dD1=dD(str(f1))t1=dD1.normalize().as_tuple()if(len(t1[1])<12):# if number == 12.34999999999999999999, dD1 = 12.35returndD1returnnumberdefABCDEF_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))forvinabc]e,p,q=[dD(str(v))forvinepq]divider=a**2+b**2ifdivider==0:print(thisName,'At least one of (a,b) must be non-zero.')returnNoneifdivider!=1:root=divider.sqrt()a,b,c=[(v/root)forvin(a,b,c)]distance_from_focus_to_directrix=a*p+b*q+cifdistance_from_focus_to_directrix==0:print(thisName,'distance_from_focus_to_directrix must be non-zero.')returnNoneX=e*eA=X*a**2-1B=X*b**2-1C=2*X*a*bD=2*p+2*X*a*cE=2*q+2*X*b*cF=X*c**2-p*p-q*qA,B,C,D,E,F=[reduce_Decimal_number(v)forvin(A,B,C,D,E,F)]ifflag: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)
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.
For a quick check select some random points on the curve:
# python codeforxin(-2,4,6):y=x**2/4print('\nFrom point ({}, {}):'.format(x,y))distance_to_focus=((x-p)**2+(y-q)**2)**.5distance_to_directrix=a*x+b*y+cs1='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
# python codepoints=((-4,0),(-3.5,-1.875),(3.5,1.875),(-1,3.75),(1,-3.75),)A,B,F=-0.9375,-1,15for(x,y)inpoints:# Verify that point is on curve.(A*x**2+B*y**2+F)and1/0# Create exception if sum != 0.distance_to_focus=((x-p)**2+(y-q)**2)**.5distance_to_directrix=a*x+b*y+ce=distance_to_focus/distance_to_directrixs1='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
# python codefour_points=pt1,pt2,pt3,pt4=(-7.5,9),(-7.5,-9),(22.5,21),(22.5,-21)for(x,y)infour_points:# Verify that point is on curve.sum=1.25*y**2-x**2-45sumand1/0# Create exception if sum != 0.distance_to_focus=((x-p)**2+(y-q)**2)**.5distance_to_directrix=a*x+b*y+ce=distance_to_focus/distance_to_directrixs1='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
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.
# python codedefsolve_quadratic(abc):'''result = solve_quadratic (abc)result may be : [] [ root1 ] [ root1, root2 ]'''a,b,c=abcifa==0:return[-c/b]disc=b**2-4*a*cifdisc<0:return[]two_a=2*aifdisc==0:return[-b/two_a]root=disc.sqrt()r1,r2=(-b-root)/two_a,(-b+root)/two_areturn[r1,r2]defcalculate_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))forvinABC]# Quadratic function in K: (a_)K**2 + (b_)K + (c_) = 0a_=4*A_*B_-C_*C_b_=4*(A_+B_)c_=4values_of_K=solve_quadratic((a_,b_,c_))ifflag: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=[]forKinvalues_of_K:A,B,C=[reduce_Decimal_number(v*K)forvin(A_,B_,C_)]X=A+B+2ifX<=0:# Here is one place where the spurious value of K may be eliminated.ifflag:print(' K = {}, X = {}, continuing.'.format(K,X))continueaa=reduce_Decimal_number((A+1)/X)ifflag:print(' K =',K)forstrxin('A','B','C','X','aa'):print(' ',strx,eval(strx))ifaa==0:a=dD(0);b=dD(1)else:a=aa.sqrt();b=C/(2*X*a)Kab=[reduce_Decimal_number(v)forvin(K,a,b)]output+=[Kab]ifflag:print(thisName)fortinrange(0,len(output)):str1=' output[{}] = {}'.format(t,output[t])print(str1)returnoutput
# python codedefcompare_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.forv1,v2inzip(ABCDEF1,ABCDEF2):status=(bool(v1)==bool(v2))ifnotstatus:print(thisName)print(' mismatch:',v1,v2)returnstatus# Results of v1/v2 must all be the same.set1={(v1/v2)for(v1,v2)inzip(ABCDEF1,ABCDEF2)ifv2}status=(len(set1)==1)ifstatus:quotient,=list(set1)else:quotient='??'L1=[];L2=[];L3=[]forminrange(0,6):bottom=ABCDEF2[m]ifnotbottom:continuetop=ABCDEF1[m]L1+=[str(top)];L3+=[str(bottom)]forminrange(0,len(L1)):L2+=[(sorted([len(v)forvin(L1[m],L3[m])]))[-1]]# maximum value.forminrange(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))returnstatusdefcalculate_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))forvinABCDEF_]ifflag:v1,v2,v3,v4,v5,v6=ABCDEFstr1='({})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)inresult:A,B,C,D,E,F=[reduce_Decimal_number(K*v)forvinABCDEF]X=A+B+2e=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*Fvalues_of_c=solve_quadratic((a_,b_,c_))# values_of_c may be empty in which case this value of K is not used.forcinvalues_of_c:p=(D-2*X*a*c)/2q=(E-2*X*b*c)/2abc=[reduce_Decimal_number(v)forvin(a,b,c)]epq=[reduce_Decimal_number(v)forvin(e,p,q)]output+=[(abc,epq)]ifflag:print(thisName)str1=' ({})x^2 + ({})y^2 + ({})xy + ({})x + ({})y + ({}) = 0'.format(A,B,C,D,E,F)print(str1)ifvalues_of_c:str1=' K = {}. values_of_c = {}'.format(K,values_of_c)else:str1=' K = {}. values_of_c = {}'.format(K,'EMPTY')print(str1)iflen(output)notin(1,2):# This should be impossible.print(thisName)print(' Internal error: len(output) =',len(output))1/0ifflag:# Check output and print results.L1=[]for((a,b,c),(e,p,q))inoutput: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_ = 0c_=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.ifnotcompare_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)iflen(set2)!=1:print(' set2 =',set2)3/0returnoutput
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
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 codeABCDEF=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
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
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
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 codedefcalculate_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))forvinABCDEF]_a,_b,_c=B,C,AX,Y=XYif_a==0:# Allow for the possibility that one asymptote may be a vertical line.m=1,0# m = 1/0temp+=[m]# _b m + _c = 0## -_c# m = ---# _b#m=-_c,_btemp+=[m]else:discr=_b**2-4*_a*_croot=discr.sqrt()top1,top2=(-_b-root),(-_b+root)bottom=2*_am=top1,bottomtemp+=[m]m=top2,bottomtemp+=[m]output=[]forp,qintemp:# p# y = - x + c# q## qy = px + qc# px - qy + qc = 0# a1 x + b1 y + c1 = 0a1,b1=p,-q# line contains mid_point. Using c = - (ax+by)c1=-(a1*X+b1*Y)abc=tuple(reduce_Decimal_number(v)forvin(a1,b1,c1))output+=[abc]ifflag:# Print output and check results.fora1,b1,c1inoutput:print('asymptote: ({})x + ({})y + ({}) = 0'.format(a1,b1,c1))ifnotb1:continuedefalmost_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=0forvininput:sum+=vifv>0:sump+=v# sump + sumn = sumsumn=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))or1/0# b_ = 2Bcm + cC + D + Em. b_ should be 0.almost_zero((2*B*c*m,c*C,D,E*m))or2/0returnoutput
"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:
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)
Length of major axis: = 26.52631578947368421052 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 codeABCDEF=A,B,C,D,E,F=1771,1204,1944,-44860,-18520,214400# ellipseresult=calculate_abc_epq(ABCDEF)(len(result)==2)or1/0# ellipse or hyperbola(abc1,epq1),(abc2,epq2)=resulta1,b1,c1=abc1;e1,p1,q1=epq1a2,b2,c2=abc2;e2,p2,q2=epq2(e1==e2)or2/0(1>e1>0)or3/0print('({})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))
# python codedefsum_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. '''globalTolerancesump=sumn=0forvininput:ifv>0:sump+=velifv<0:sumn-=vsum=sump-sumnifabs(sum)<Tolerance:return(type(Tolerance))(0)min,max=sorted((sumn,sump))ifabs(sum)<=Tolerance*min:return(type(Tolerance))(0)returnsum
# axis is perpendicular to directrix.ax,bx=b1,-a1# axis contains foci. ax + by + c = 0cx=reduce_Decimal_number(-(ax*p1+bx*q1))axis=ax,bx,cxprint(' Axis : ({})x + ({})y + ({}) = 0'.format(ax,bx,cx))print(' Eccentricity = {}'.format(e1))print()print(' Directrix 1 : ({})x + ({})y + ({}) = 0'.format(a1,b1,c1))F1=p1,q1# Focus 1.print(' F1 : ({}, {})'.format(p1,q1))distance_from_F1_to_ID1=abs(a1*p1+b1*q1+c1)# Direction cosines along axis from F1 towards ID1:dx,dy=a1,b1# Intercept on directrix1ID1=xID1,yID1=p1+dx*distance_from_F1_to_ID1,q1+dy*distance_from_F1_to_ID1# Check. ID1 should be on Directrix 1. (a1)xID1 + (b1)yID1 + (c1) should be 0.values=a1*xID1,b1*yID1,c1ifsum_zero(values):# A little trial and error here to get direction of dx,dy correct.dx,dy=-a1,-b1ID1=xID1,yID1=p1+dx*distance_from_F1_to_ID1,q1+dy*distance_from_F1_to_ID1values=a1*xID1,b1*yID1,c1ifsum_zero(values):# This should not happen.print('Internal error 1.')1/0print(' Intercept ID1 : ({}, {})'.format(xID1,yID1))# Intercept1 on curve# # distance_from_F1_to_curve e2# ------------------------- = ------# distance_from_F1_to_ID1 1 + e2# distance_from_F1_to_curve=distance_from_F1_to_ID1*e2/(1+e2)xI1,yI1=p1+dx*distance_from_F1_to_curve,q1+dy*distance_from_F1_to_curveI1=xI1,yI1=[reduce_Decimal_number(v)forvin(xI1,yI1)]print(' Intercept I1 : ({}, {})'.format(xI1,yI1))
# 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=e1*distance_from_F1_to_ID1# latus rectum 1# Focal chord has equation (afc)x + (bfc)y + (cfc) = 0.afc,bfc=a1,b1cfc=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_rectumprint(' Point P : ({}, {})'.format(xP,yP))U=xU,yU=p1-dlx*half_latus_rectum,q1-dly*half_latus_rectumprint(' Point U : ({}, {})'.format(xU,yU))distance=((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
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'),)fornameint1:value=eval(name)x,y=[reduce_Decimal_number(v)forvinvalue]print('{} : ({}, {})'.format((name+' ')[:2],x,y))values=A*x**2,B*y**2,C*x*y,D*x,E*y,Fsum_zero(values)and3/0# Relative to Directrix 1 and Focus 1:distance_to_F1=((x-p1)**2+(y-q1)**2).sqrt()distance_to_directrix1=a1*x+b1*y+c1e1=distance_to_F1/distance_to_directrix1print(' 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+c2e2=distance_to_F2/distance_to_directrix2e2=reduce_Decimal_number(e2)print(' e2 =',e2)# Clean value is printed.
Note the differences between "raw" values of and "clean" values of