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.

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 ${\displaystyle \sin \alpha }$),

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

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 ${\displaystyle a<b}$ 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: ${\displaystyle x=\pm {\frac {a}{e}}}$

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)

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 ${\displaystyle y=\pm {\frac {b}{a}}x}$

In cartesian geometry in two dimensions hyperbola is locus of a point ${\displaystyle P}$ that moves relative to two fixed points called ${\displaystyle foci}$${\displaystyle :F_{1},F_{2}.}$ The distance ${\displaystyle F_{1}F_{2}}$ from one ${\displaystyle focus\ (F_{1})}$ to the other ${\displaystyle focus\ (F_{2})}$ is non-zero. The absolute difference of the distances ${\displaystyle (PF_{1},PF_{2})}$ from point to foci is constant. ${\displaystyle PF_{1}-PF_{2}=K.}$ See figure 1. Center of hyperbola is located at the origin ${\displaystyle O(0,0)}$ and the foci ${\displaystyle (F_{1},F_{2})}$ are on the ${\displaystyle X\ axis}$ at distance ${\displaystyle c}$ from ${\displaystyle O.}$ ${\displaystyle F_{1}}$ has coordinates ${\displaystyle (-c,0).F_{2}}$ has coordinates ${\displaystyle (c,0)}$. Line segments ${\displaystyle OF_{1}=OF_{2}=c.}$ Each point ${\displaystyle (V_{1},V_{2})}$ where the curve intersects the transverse axis is called a ${\displaystyle vertex.\ V_{1},V_{2}}$ are the vertices of the ellipse. By definition ${\displaystyle PF_{1}-PF_{2}=V_{2}F_{1}-V_{2}F_{2}=V_{1}F_{2}-V_{1}F_{1}=K.}$ ${\displaystyle \therefore V_{2}F_{1}-V_{2}F_{2}=V_{2}F_{1}-V_{1}F_{1}=V_{1}V_{2}=K=2a,}$ the length of the ${\displaystyle transverse\ axis\ (V_{1}V_{2}).}$ ${\displaystyle OV_{1}=OV_{2}=a.}$

${\displaystyle PF_{1}-PF_{2}=2a}$

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

${\displaystyle PF_{1}={\sqrt {(x-(-c))^{2}+(y-0)^{2}}}={\sqrt {(x+c)^{2}+y^{2}}}=m}$

${\displaystyle PF_{2}={\sqrt {(x-c)^{2}+y^{2}}}=n}$

${\displaystyle m^{2}=M=(x+c)^{2}+y^{2};\ n^{2}=N=(x-c)^{2}+y^{2}.}$

${\displaystyle {\sqrt {(x+c)^{2}+y^{2}}}-{\sqrt {(x-c)^{2}+y^{2}}}=2a}$

${\displaystyle m-n=2a}$

${\displaystyle m^{2}-2mn+n^{2}=M-2mn+N=4aa}$

${\displaystyle -2mn=4aa-M-N}$

${\displaystyle 4m^{2}n^{2}=(4aa-M-N)^{2}}$

${\displaystyle 4MN-(4aa-M-N)^{2}=0}$

Make appropriate substitutions, expand and result is:

${\displaystyle 16aaxx-16ccxx+16aayy-16aaaa+16aacc=0}$

Simplify, gather like terms and recall that, for hyperbola, ${\displaystyle c>a.}$

${\displaystyle (aa-cc)xx+aayy-aa(aa-cc)=0}$

${\displaystyle (cc-aa)xx-aayy-aa(cc-aa)=0}$

Let ${\displaystyle b^{2}=c^{2}-a^{2}}$

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

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

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

Curve in diagram has:

${\displaystyle a=5,c=6.25}$

${\displaystyle b^{2}=c^{2}-a^{2}=14.0625}$

equation ${\displaystyle (14.0625)x^{2}-(25)y^{2}=351.5625}$ or ${\displaystyle {\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}={\frac {x^{2}}{5^{2}}}-{\frac {y^{2}}{3.75^{2}}}=1}$

where ${\displaystyle a^{2}=5^{2},b^{2}=3.75^{2}.}$

${\displaystyle PF_{1}-PF_{2}=2a}$

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

Let Focus 1, ${\displaystyle F_{1}}$ have coordinates ${\displaystyle (0,-c).}$

Let Focus 2, ${\displaystyle F_{2}}$ have coordinates ${\displaystyle (0,c).}$

${\displaystyle PF_{1}={\sqrt {(x)^{2}+(y-(-c))^{2}}}={\sqrt {x^{2}+(y+c)^{2}}}=m}$

${\displaystyle PF_{2}={\sqrt {x^{2}+(y-c)^{2}}}=n}$

${\displaystyle m^{2}=M=x^{2}+(y+c)^{2};\ n^{2}=N=x^{2}+(y-c)^{2}.}$

Make substitutions as above, expand and result is:

${\displaystyle aayy-ccyy+aaxx+aacc-aaaa=0}$

Simplify, gather like terms and recall that, for hyperbola, ${\displaystyle c>a.}$

${\displaystyle (aa-cc)yy+aaxx+aa(cc-aa)=0}$

${\displaystyle (cc-aa)yy-aaxx-aa(cc-aa)=0}$

Let ${\displaystyle b^{2}=c^{2}-a^{2}}$

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

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

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

Curve in diagram has:

${\displaystyle a=4,c=5}$

${\displaystyle b^{2}=5^{2}-4^{2}=3^{2}=9}$

equation ${\displaystyle (9)y^{2}-(16)x^{2}=(9)(16)}$ or ${\displaystyle {\frac {y^{2}}{a^{2}}}-{\frac {x^{2}}{b^{2}}}={\frac {y^{2}}{4^{2}}}-{\frac {x^{2}}{3^{2}}}=1}$

where ${\displaystyle a^{2}=4^{2},b^{2}=3^{2}.}$

With transverse axis oblique the two foci are defined as:

${\displaystyle F_{1}:\ (p,q)}$

${\displaystyle F_{2}:\ (-p,-q)}$

where both ${\displaystyle (p,q)}$ are non-zero.

Distance from center to focus ${\displaystyle =c={\sqrt {p^{2}+q^{2}}},}$ and ${\displaystyle c>a.}$

Let ${\displaystyle m={\sqrt {(x-p)^{2}+(y-q)^{2}}}}$

Let ${\displaystyle n={\sqrt {(x-(-p))^{2}+(y-(-q))^{2}}}}$

By definition: ${\displaystyle m-n=2a}$

If you make the substitutions as before, result is:

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

${\displaystyle A=a^{2}-p^{2}}$

${\displaystyle B=a^{2}-q^{2}}$

${\displaystyle C=-2pq}$

${\displaystyle F=a^{2}(p^{2}+q^{2}-a^{2})}$

Curve in diagram has:

• Foci at (24,7), (-24,-7)

• a = 20

• equation ${\displaystyle (-176)x^{2}+(351)y^{2}+(-336)xy+(90000)=0.}$

# 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

The expression "reversing the process" means calculating the values of ${\displaystyle a,p,q}$ when given equation of hyperbola at origin, the familiar values ${\displaystyle A,B,C,F.}$

Consider the equation of a hyperbola at origin: ${\displaystyle 351x^{2}-176y^{2}+336xy+90000=0.}$ This is a hyperbola where ${\displaystyle A,B,C,F=351,-176,336,90000.}$

This hyperbola may be expressed as ${\displaystyle 35.1x^{2}-17.6y^{2}+33.6xy+9000.0=0}$ or ${\displaystyle 3510x^{2}-1760y^{2}+3360xy+900000=0}$ or ${\displaystyle (K)351x^{2}-(K)176y^{2}+(K)336xy+(K)90000=0}$ where ${\displaystyle K}$ is any real, non-zero number.

However, when this hyperbola is expressed as ${\displaystyle 351x^{2}-176y^{2}+336xy+90000=0,}$ this format is the hyperbola expressed in "standard form," a notation that greatly simplifies the calculation of ${\displaystyle a,p,q.}$

Modify the equations for ${\displaystyle A,B,C,F}$ slightly:

${\displaystyle KA=a^{2}-p^{2}}$

${\displaystyle KB=a^{2}-q^{2}}$

${\displaystyle KC=-2pq}$

${\displaystyle KF=a^{2}(p^{2}+q^{2}-a^{2})}$

There are four simultaneous equations with four known values ${\displaystyle A,B,C,F}$ and four unknown: ${\displaystyle K,a,p,q.}$

${\displaystyle K={\frac {4F}{C^{2}-4AB}}.}$

For ${\displaystyle A,B,C,F=35.1,-17.6,33.6,9000.0,\ K=10.}$

For ${\displaystyle A,B,C,F=351,-176,336,90000,\ K=1,}$ the ideal condition.

# 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)

Hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. Let ${\displaystyle {\frac {c}{a}}=e}$ where: ${\displaystyle a}$ is non-zero, ${\displaystyle c>a,}$ ${\displaystyle c=a+v.}$ Therefore, ${\displaystyle e>1.}$ Let directrix have equation ${\displaystyle x=d}$ where ${\displaystyle {\frac {a}{d}}=e.}$ At vertex ${\displaystyle V_{2}:}$ ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}}$ ${\displaystyle ={\frac {v}{u}}={\frac {c-a}{a-d}}}$ ${\displaystyle ={\frac {ae-a}{a-a/e}}={\frac {ae-a}{(ae-a)/e}}}$ ${\displaystyle ={\frac {(ae-a)e}{ae-a}}=e.}$ At point ${\displaystyle P:}$ ${\displaystyle b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}}$ ${\displaystyle a^{2}y^{2}=b^{2}x^{2}-a^{2}b^{2}}$ ${\displaystyle y^{2}={\frac {b^{2}x^{2}-a^{2}b^{2}}{a^{2}}}}$ ${\displaystyle ={\frac {(c^{2}-a^{2})x^{2}-a^{2}(c^{2}-a^{2})}{a^{2}}}}$ ${\displaystyle ={\frac {(a^{2}e^{2}-a^{2})x^{2}-a^{2}(c^{2}-a^{2})}{a^{2}}}}$ ${\displaystyle =(e^{2}-1)x^{2}-(a^{2}e^{2}-a^{2})}$ ${\displaystyle =e^{2}x^{2}-x^{2}-a^{2}e^{2}+a^{2}}$ At point ${\displaystyle P_{1}\ x=c:}$ ${\displaystyle y^{2}=e^{2}c^{2}-c^{2}-a^{2}e^{2}+a^{2}}$ ${\displaystyle =e^{2}a^{2}e^{2}-a^{2}e^{2}-a^{2}e^{2}+a^{2}}$ ${\displaystyle =a^{2}(e^{4}-2e^{2}+1)}$ ${\displaystyle =a^{2}(e^{2}-1)^{2}}$ ${\displaystyle y=a(e^{2}-1).}$ ${\displaystyle {\text{distance to directrix}}=c-d}$ ${\displaystyle =ae-a/e}$ ${\displaystyle =(aee-a)/e}$ ${\displaystyle =a(e^{2}-1)/e}$ ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}}$ ${\displaystyle ={\frac {a(e^{2}-1)}{a(e^{2}-1)/e}}}$ ${\displaystyle ={\frac {a(e^{2}-1)e}{a(e^{2}-1)}}=e.}$ Aim of this section is to prove that : ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}=e\ \dots \ (3)}$ Statement ${\displaystyle (3)}$ has been proved for two specific points, vertex ${\displaystyle V_{2}}$ and point ${\displaystyle P_{1}.}$ Section under "Proof" below proves that statement (3) is true for any point ${\displaystyle P}$ on hyperbola.

As expressed above in statement ${\displaystyle 3,}$ second definition of hyperbola states that hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. This section proves that this definition is true for any point ${\displaystyle P}$ on hyperbola. At point ${\displaystyle P:}$ ${\displaystyle y^{2}=e^{2}x^{2}-x^{2}+a^{2}-a^{2}e^{2}}$ base ${\displaystyle =x-c=x-ae}$ ${\displaystyle ({\text{distance}}\ F_{2}P)^{2}=y^{2}+{\text{base}}^{2}=y^{2}+(x-ae)^{2}}$ ${\displaystyle =a^{2}-2aex+e^{2}x^{2}}$ ${\displaystyle =(ex-a)^{2}}$ ${\displaystyle {\text{distance to Focus2}}={\text{distance}}\ F_{2}P=ex-a}$ ${\displaystyle {\text{distance to Directrix2}}=x-d=x-{\frac {a}{e}}={\frac {ex-a}{e}}}$ ${\displaystyle {\frac {\text{distance to Focus2}}{\text{distance to Directrix2}}}}$ ${\displaystyle =(ex-a){\frac {e}{(ex-a)}}}$ ${\displaystyle =e}$ Similar calculations can be used to prove the case for Focus1 ${\displaystyle (-c,0)}$ and Directrix1 ${\displaystyle (x=-d)}$ in which case: ${\displaystyle {\frac {\text{distance to Focus1}}{\text{distance to Directrix1}}}}$ ${\displaystyle =e}$ Therefore: ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}=e}$ where ${\displaystyle e>1.}$ Second definition of hyperbola has been proven valid: Hyperbola is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant, called eccentricity ${\displaystyle e.}$

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

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

# python code

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

# python code

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

To calculate the general equation three values must be known:

• Focus1 ${\displaystyle (F_{1})}$

• Focus2 ${\displaystyle (F_{2})}$

• Length of transverse axis ${\displaystyle (a*2)}$ or two_a.

Calculate center of hyperbola ${\displaystyle (G,H)}$

Calculate ${\displaystyle a,(p,q).}$

Calculate equation of hyperbola at origin ${\displaystyle Ax^{2}+By^{2}+Cxy+F=0.}$

Move hyperbola from origin to point ${\displaystyle (G,H).}$

# 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

Curve in Figure 1 below is defined by:

• Focus1, ${\displaystyle F_{1}=(-9,6)}$

• Focus2, ${\displaystyle F_{2}=(-9,102)}$

• Constant, ${\displaystyle 2a=60}$

Curve has equation: ${\displaystyle (900)x^{2}+(-1404)y^{2}+(0)xy+(16200)x+(151632)y+(-2757564)=0.}$

Curve in Figure 2 below is defined by:

• Focus1, ${\displaystyle F_{1}=(-7,24)}$

• Focus2, ${\displaystyle F_{2}=(7,-24)}$

• Constant, ${\displaystyle 2a=40}$

Curve has equation: ${\displaystyle (351)x^{2}+(-176)y^{2}+(336)xy+(0)x+(0)y+(90000)=0}$

Curve in Figure 3 below is defined by:

• Focus1, ${\displaystyle F_{1}=(13,-27)}$

• Focus2, ${\displaystyle F_{2}=(43,-11)}$

• Constant, ${\displaystyle 2a=27.2}$

Curve has equation: ${\displaystyle (-40.04)x^{2}+(120.96)y^{2}+(-240)xy+(-2317.76)x+(11316.48)y+(159198.4384)=0}$

The expression "reversing the process" means calculating the values of ${\displaystyle F_{1},F_{2}}$ and length of transverse axis when given equation of general hyperbola, the familiar values ${\displaystyle A,B,C,D,E,F.}$

Calculate center of hyperbola.

Move hyperbola to origin.

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

Calculate ${\displaystyle F_{1},F_{2}}$ and length of transverse axis, ${\displaystyle (2a).}$

# 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

A hyperbola has equation: ${\displaystyle f(x,y)=176x^{2}-351y^{2}-336xy-2528x+6504y-89104=0.}$ Calculate two foci and length of transverse axis. # python code ABCDEF = 176, -351, -336, -2528, 6504, -89104 two_a, F1, F2 = hyperbola_2a_F1_F2 (ABCDEF) print ('Length of transverse axis =',two_a ) print ('F1 = ({},{})'.format(F1[0], F1[1] )) print ('F2 = ({},{})'.format(F2[0], F2[1] )) Length of transverse axis = 40 F1 = (35,-3) F2 = (-13,11) To check calculation and for more information about hyperbola: # python code hyperbola_2a_F1_F2 (ABCDEF, 2) # Set flag. hyperbola_2a_F1_F2 (ABCDEF, flag = 2) : (176.00)x^2 + (-351.00)y^2 + (-336)xy + (-2528.00)x + (6504.00)y + (-89104.0000) = 0 (-176.00)x^2 + (351.00)y^2 + (336.0)xy + (2528.000)x + (-6504.00)y + (89104.0000) = 0 Standard form. (176.00)x^2 + (-351.00)y^2 + (-336)xy + (-90000.0000) = 0 At origin. K = -1 G,H = (Decimal('11'), Decimal('4')) a = 20.0 2*a = 40.0 two_c = 50.0 e = 1.25 F1: (35.0, -3) (x - (35.0))^2 + (y - (-3))^2 = 1 F2: (-13.0, 11) (x - (-13.0))^2 + (y - (11))^2 = 1 f1: (24.0, -7) (x - (24.0))^2 + (y - (-7))^2 = 1 f2: (-24.0, 7) (x - (-24.0))^2 + (y - (7))^2 = 1 axis: (-14)x + (-48.0)y + (346.0) = 0