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
edit
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
editS 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
editA 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
editA 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
 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.
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Equation of hyperbola at origin
editTransverse axis horizontal
edit 
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
edit
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
edit
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
edit
# 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
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Examples
edit
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Reversing the process
editThe 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
edit
# 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)
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Second definition of hyperbola
edit
 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.
Therefore, Let directrix have equation where
Statement has been proved for two specific points, vertex and point
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Proof
edit|
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.
base
Therefore: where
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
editIn 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
editTo 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
editCurve 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:
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Reversing the process
editThe 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
# 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
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Example
edit
 Equation of hyperbola is given. This section calculates length of transverse axis, A hyperbola has equation: Calculate two foci and # 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
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Asymptotes of hyperbola
edit
 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:
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
Hyperbola has equation Calculate point of intersection of
Substitute for in This does not make sense. There is no real point of intersection of
Calculate point of intersection of
Provided that are non-zero, is a real number.
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Line and hyperbola
editThis section describes possibilities that arise when we consider intersection of line and hyperbola.
At origin
editLet 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
editLet 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
# 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
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Examples
editWith no common point
edit
 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
edit
 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
edit
 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
edit
# 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
|
Gallery
edit
|
Family of hyperbolas
edit
 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.
|
Rectangular Hyperbolas
editRectangular Hyperbolas are special cases of hyperbolas where the asymptotes are perpendicular. These have the general equation:
 A rectangular hyperbola has eccentricity If hyperbola is expressed as then:
|
Details
edit
 Diagram 2 shows detail of quadrant 1 of Diagram 1.
|
Examples
editRectangular hyperbola at origin
edit
 Curve in Figure 1 has equation:
Vertex has coordinates Vertex has coordinates
Focus has coordinates Focus has coordinates
Point has coordinates Point has coordinates Directrix1, directrix through point has equation Directrix2, directrix through point has equation
Asymptote2 has equation: Slope of asymptote1: Slope of asymptote2: Product of slopes: Asymptotes are perpendicular. |
Rectangular hyperbola parallel to Y axis
edit
 Curve is moved from origin to point Curve in Figure 2 has equation:
Vertex has coordinates Vertex has coordinates
Focus has coordinates Focus has coordinates
Point has coordinates Point has coordinates Directrix1, directrix through point has equation Directrix2, directrix through point has equation
Asymptote2 has equation: Slope of asymptote1: Slope of asymptote2: Product of slopes: Asymptotes are perpendicular. |
Rectangular hyperbola, random
edit
 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:
Vertex has coordinates Vertex has coordinates
Focus has coordinates Focus has coordinates
Point has coordinates Point has coordinates Conjugate axis through point has equation Directrix1, directrix through point has equation Directrix2, directrix through point has equation
Asymptote2 has equation: Slope of asymptote1: Slope of asymptote2: Product of slopes: Asymptotes are perpendicular. |
Conic sections generally
editWithin 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
editThe 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:
For example, directrix and directrix produce same result. |
Implementation
edit# 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
editParabola
editEvery parabola has eccentricity
 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
|
Gallery
edit|
Curve in Figure 1 below has:
Curve in Figure 2 below has:
Curve in Figure 3 below has:
|
