# Ellipse

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## Ellipse

Figure 1: Ellipse (red curve) at origin with major axis horizontal.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Foci are points ${\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}$
Line segment ${\displaystyle V_{1}OV_{2}}$ is the ${\displaystyle major\ axis.}$
${\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=B_{1}F_{1}=B_{1}F_{2}=a.}$
Line segment ${\displaystyle B_{1}OB_{2}}$ is the ${\displaystyle minor\ axis.\ OB_{1}=OB_{2}=b.}$
Each line ${\displaystyle L_{1}F_{1}R_{1},L_{2}F_{2}R_{2}}$ is a ${\displaystyle latus\ rectum.}$
Each line ${\displaystyle x=D_{1},x=D_{2}}$ is a ${\displaystyle directrix.}$
${\displaystyle PF_{1}+PF_{2}=2a.}$

In cartesian geometry in two dimensions the ellipse is the 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 sum of the distances ${\displaystyle (PF_{1},PF_{2})}$ from point to foci is constant.

${\displaystyle PF_{1}+PF_{2}=K.}$ See figure 1.

The center of the ellipse 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 ${\displaystyle OF_{1}=OF_{2}=c.}$

By definition ${\displaystyle PF_{1}+PF_{2}=B_{1}F_{1}+B_{1}F_{2}=V_{1}F_{1}+V_{1}F_{2}=K.}$

${\displaystyle V_{1}F_{1}=V_{2}F_{2}.\ \therefore V_{1}F_{1}+V_{1}F_{2}=V_{1}F_{2}+V_{2}F_{2}=V_{1}V_{2}=K=2a,}$ the length of the ${\displaystyle major\ axis\ (V_{1}V_{2}).\ OV_{1}=OV_{2}=a.}$

Each point ${\displaystyle (V_{1},V_{2})}$ where the curve intersects the major axis is called a ${\displaystyle vertex.\ V_{1},V_{2}}$ are the vertices of the ellipse.

Line segment ${\displaystyle B_{1}B_{2}}$ perpendicular to the major axis at the midpoint of the major axis is the ${\displaystyle minor\ axis}$ with length ${\displaystyle 2b.\ OB_{1}=OB_{2}=b.}$

${\displaystyle B_{1}F_{1}+B_{1}F_{2}=2a;\ B_{1}F_{1}=B_{1}F_{2}=a.\ \therefore a^{2}=b^{2}+c^{2}.}$

Any line segment that intersects the curve in two places is a ${\displaystyle chord.}$ A chord through the focus is a ${\displaystyle focal\ chord.}$ Each focal chord ${\displaystyle L_{1}F_{1}R_{1},\ L_{2}F_{2}R_{2}}$ perpendicular to the major axis is a ${\displaystyle latus\ rectum.}$

Equation of the ellipse

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

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

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

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

${\displaystyle MM=(x+c)^{2}+y^{2};\ NN=(x-c)^{2}+y^{2}.}$

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

${\displaystyle M+N=2a}$

${\displaystyle MM+2MN+NN=4aa}$

${\displaystyle 2MN=4aa-MM-NN}$

${\displaystyle 4MMNN=(4aa-MM-NN)^{2}}$

${\displaystyle 4MMNN-(4aa-MM-NN)^{2}=0}$

Make appropriate substitutions, expand and the result is:

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

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

${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$ or ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$

If the equation ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$ is expressed as ${\displaystyle Ax^{2}+By^{2}+C=0:}$

${\displaystyle a^{2}={\frac {-C}{A}};\ b^{2}={\frac {-C}{B}}.}$

Length of latus rectum

${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$

${\displaystyle b^{2}c^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$

${\displaystyle b^{2}(a^{2}-b^{2})+a^{2}y^{2}-a^{2}b^{2}=0}$

${\displaystyle b^{2}a^{2}-b^{4}+a^{2}y^{2}-a^{2}b^{2}=0}$

${\displaystyle a^{2}y^{2}=b^{4}}$

${\displaystyle y^{2}={\frac {b^{4}}{a^{2}}}}$

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

Length of latus rectum ${\displaystyle =L_{1}R_{1}=L_{2}R_{2}={\frac {2b^{2}}{a}}.}$

## The directrices

Figure 1: Ellipse (red curve) at origin with major axis horizontal.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Foci are points ${\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}$
Directrices are lines ${\displaystyle x=D_{1},\ x=D_{2},\ D_{1}=-D_{2}.}$ ${\displaystyle OV_{1}=OV_{2}=a.}$
${\displaystyle eccentricity\ e={\frac {c}{a}}={\frac {F_{2}V_{2}}{V_{2}D_{2}}}={\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {OV_{2}}{OD_{2}}}}$
${\displaystyle e={\frac {PF_{1}}{PE_{1}}}}$ (yellow lines); ${\displaystyle e={\frac {PF_{2}}{PE_{2}}}}$ (purple lines)
To define ellipse specify ${\displaystyle focus,\ directrix,\ e.}$

See Figure 1. The vertical line through ${\displaystyle D_{2}}$ with equation ${\displaystyle x=D_{2}}$ is a ${\displaystyle directrix}$ of the ellipse. Likewise for the vertical line ${\displaystyle x=D_{1};\ D_{1}=-D_{2}.}$

A second definition of the ellipse: the locus of a point that moves relative to a point, the focus, and a fixed line called the directrix, so that the distance from point to focus and the distance from point to line form a constant ratio ${\displaystyle e=eccentricity,\ 0

Consider the point ${\displaystyle V_{2}}$ in the figure. ${\displaystyle {\frac {F_{2}V_{2}}{V_{2}D_{2}}}=e.}$

Let the length ${\displaystyle V_{2}D_{2}=d.\ {\frac {a-c}{d}}=e.\ d={\frac {a-c}{e}}}$

Consider the point ${\displaystyle V_{1}}$:

${\displaystyle {\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {a+c}{2a+d}}=e=(a+c)/(2a+{\frac {a-c}{e}})=(a+c)/({\frac {2ae+a-c}{e}})={\frac {e(a+c)}{2ae+a-c}}.}$

${\displaystyle 2ae+a-c=a+c;\ 2ae=2c.}$

${\displaystyle e={\frac {c}{a}}.}$

Length ${\displaystyle OD_{2}=a+d=a+{\frac {a-c}{e}}=a+{\frac {a-c}{c/a}}=a+{\frac {a(a-c)}{c}}={\frac {ac+aa-ac}{c}}={\frac {a^{2}}{c}}}$.

Distance from center to directrix ${\displaystyle =OD_{2}={\frac {a^{2}}{c}}={\frac {a^{2}}{ea}}={\frac {a}{e}}={\frac {c}{e^{2}}}.\ e={\frac {OV_{2}}{OD_{2}}}.}$

Distance from center to directrix = ${\displaystyle OF_{2}+F_{2}D_{2}=c+d_{0}={\frac {c}{e^{2}}}}$.

${\displaystyle e^{2}c+e^{2}d_{0}=c;\ c-e^{2}c=e^{2}d_{0};\ c={\frac {e^{2}d_{0}}{1-e^{2}}}}$.

${\displaystyle 1-e^{2}=1-{\frac {c^{2}}{a^{2}}}={\frac {a^{2}-c^{2}}{a^{2}}}={\frac {b^{2}}{a^{2}}}}$.

${\displaystyle {\frac {e^{2}}{1-e^{2}}}={\frac {c^{2}}{a^{2}}}/{\frac {b^{2}}{a^{2}}}={\frac {c^{2}}{b^{2}}}}$.

Distance from focus to directrix ${\displaystyle =F_{2}D_{2}={\frac {b^{2}}{c}}}$.

Ellipse using focus and directrix

Figure 2: Ellipse (red curve) at origin with major axis vertical.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Focus at point ${\displaystyle F}$${\displaystyle :(0,-c).}$
Directrix is line ${\displaystyle DE}$${\displaystyle :\ y={\frac {-c}{e^{2}}}.}$
Point ${\displaystyle P}$ has coordinates: ${\displaystyle (x,y).}$
${\displaystyle eccentricity\ e={\frac {PF}{PE}}}$ (yellow lines)

See Figure 2. Let the point ${\displaystyle P}$ be ${\displaystyle (x,y)}$, the focus ${\displaystyle F}$ be ${\displaystyle (0,-c)}$ and the directrix ${\displaystyle DE}$ have equation ${\displaystyle y={\frac {-c}{e^{2}}}}$ where ${\displaystyle 1>e>0.}$ The directrix is horizontal and the major axis vertical through the origin ${\displaystyle (0,0)}$.

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

${\displaystyle PE=y+{\frac {c}{e^{2}}}}$

${\displaystyle e={\frac {PF}{PE}};\ PF=ePE.}$

${\displaystyle {\sqrt {x^{2}+(y+c)^{2}}}=e(y+{\frac {c}{e^{2}}})}$

${\displaystyle e{\sqrt {x^{2}+(y+c)^{2}}}=ee(y+{\frac {c}{e^{2}}})=eey+c}$

${\displaystyle ee(x^{2}+(y+c)^{2})=(eey+c)^{2}}$

${\displaystyle ee(x^{2}+(y+c)^{2})-(eey+c)^{2}=0}$

Expand and the result is:

${\displaystyle +eexx+eeyy-eeeeyy+ccee-cc=0}$

${\displaystyle xx+yy-eeyy+cc-{\frac {cc}{ee}}=0}$

${\displaystyle xx+(1-ee)yy+cc-aa=0}$

${\displaystyle xx+(1-ee)yy-(aa-cc)=0}$

${\displaystyle xx+(1-ee)yy-bb=0}$

${\displaystyle 1-ee=1-{\frac {cc}{aa}}={\frac {aa-cc}{aa}}={\frac {bb}{aa}}}$

${\displaystyle xx+({\frac {bb}{aa}})yy-bb=0}$

${\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0}$

Compare this equation with the equation generated earlier: ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$.

When the equation is ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0,}$ the major axis is horizontal.

When the equation is ${\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0,}$ the major axis is vertical.

## General ellipse at origin

Figure 1: Ellipse (red curve) at origin with major axis oblique.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Foci are points ${\displaystyle F_{1}(-p,-q),\ F_{2}(p,q).}$
${\displaystyle OF_{1}=OF_{2}=c={\sqrt {p^{2}+q^{2}}}.}$
Line segment ${\displaystyle V_{1}OV_{2}}$ is the ${\displaystyle major\ axis.}$
${\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=a.}$
Line ${\displaystyle D_{1}OD_{2}}$ is the major axis extended.
Lines perpendicular to ${\displaystyle D_{1}OD_{2}}$ at ${\displaystyle D_{1},F_{1},O,F_{2},D_{2}}$ are directrix, latus rectum. minor axis, latus rectum, directrix.
${\displaystyle OD_{1}=OD_{2}={\frac {a^{2}}{c}}.}$
${\displaystyle F_{1}D_{1}=F_{2}D_{2}={\frac {b^{2}}{c}}.}$
${\displaystyle PF_{1}+PF_{2}=2a.}$

The general ellipse allows for the major axis to have slope other than horizontal or vertical. See Figure 1.

The line ${\displaystyle V_{1}V_{2}}$ is the major axis of the ellipse shown in red. Points ${\displaystyle F_{1},F_{2}}$ are the foci with coordinates ${\displaystyle (-p,-q),(p,q)}$ respectively.

Point ${\displaystyle P(x,y)}$ is any point on the curve. By definition ${\displaystyle PF_{1}+PF_{2}=2a.}$

Length ${\displaystyle PF_{1}={\sqrt {(x-(-p))^{2}+(y-(-q))^{2}}}={\sqrt {(x+p)^{2}+(y+q)^{2}}}=M.}$

Length ${\displaystyle PF_{2}={\sqrt {(x-p)^{2}+(y-q)^{2}}}=N.}$

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

${\displaystyle M+N=2a}$

${\displaystyle MM+2MN+NN=4aa}$

${\displaystyle 2MN=4aa-MM-NN}$

${\displaystyle 4MMNN=(4aa-MM-NN)^{2}}$

${\displaystyle 4MMNN-(4aa-MM-NN)^{2}=0.}$

Make appropriate substitutions, expand and the result is:

${\displaystyle (aa-pp)x^{2}-2pqxy+(aa-qq)y^{2}+aapp+aaqq-aaaa=0}$.

This equation has the form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ where:

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

${\displaystyle B=-2pq}$

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

${\displaystyle D=E=0}$ because center is at origin,

${\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}=a^{2}(p^{2}+q^{2}-a^{2})=a^{2}(c^{2}-a^{2})=-a^{2}b^{2}.}$

An example

Let ${\displaystyle (p,q)}$ be (3,4) and ${\displaystyle a=8.}$

The ellipse is: ${\displaystyle 880x^{2}-384xy+768y^{2}+(0)x+(0)y-39936=0,}$ or ${\displaystyle 55x^{2}-24xy+48y^{2}+(0)x+(0)y-2496=0.}$

Reverse-engineering ellipse at origin

Given an ellipse in format ${\displaystyle Ax^{2}+Bxy+Cy^{2}+(0)x+(0)y+F=0,}$ calculate ${\displaystyle p,q,a.\ (B}$ is non-zero.${\displaystyle )}$

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

${\displaystyle B=-2pq}$

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

${\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}$

Coefficients provided could be, for example, ${\displaystyle (55,-24,48,0,0,-2496)}$ or ${\displaystyle (110,-48,96,0,0,-4992)}$

or ${\displaystyle (55k,-24k,48k,0,0,-2496k)}$ where ${\displaystyle k}$ is an arbitrary constant and all groups of coefficients define the same ellipse.

To produce consistent, correct values for ${\displaystyle p,q,a}$ the equations become:

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

${\displaystyle KB=-2pq}$

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

${\displaystyle KF=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}$

or:

${\displaystyle KA-(a^{2}-p^{2})=0}$

${\displaystyle KB-(-2pq)=0}$

${\displaystyle KC-(a^{2}-q^{2})=0}$

${\displaystyle KF-(a^{2}p^{2}+a^{2}q^{2}-a^{4})=0.}$

Solutions are:

${\displaystyle K={\frac {4F}{B^{2}-4AC}}}$. This formula for ${\displaystyle K}$ is valid if both ${\displaystyle D,E}$ are ${\displaystyle 0}$.

${\displaystyle 4(pp)^{2}+4K(A-C)pp-B^{2}K^{2}=0}$ where ${\displaystyle pp=p^{2}.}$

You should see one positive value ${\displaystyle (p^{2})}$ and one negative value ${\displaystyle (-(q^{2}))}$ for ${\displaystyle pp.}$ Choose the positive value and ${\displaystyle p={\sqrt {pp}}}$.

${\displaystyle q={\frac {-KB}{2p}}}$

${\displaystyle a={\sqrt {AK+p^{2}}}}$

The solutions become simpler if K == 1. if ( K != 1 ) { A ← KA; B ← KB; C ← KC; } and the solutions for ${\displaystyle p,q,a}$ become:

${\displaystyle 4(pp)^{2}+4(A-C)pp-B^{2}=0}$ where ${\displaystyle pp=p^{2}.}$

${\displaystyle q={\frac {-B}{2p}}}$

${\displaystyle a={\sqrt {A+p^{2}}}}$

With values ${\displaystyle p,q,a}$ available all the familiar values of the ellipse may be calculated:

${\displaystyle c={\sqrt {p^{2}+q^{2}}}}$

Equation of major axis: ${\displaystyle y={\frac {q}{p}}x;\ qx-py+0=0;\ {\frac {q}{c}}x-{\frac {p}{c}}y+0=0}$ in normal form.

Equation of minor axis: ${\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y+0=0}$ in normal form.

Equations of directrices: ${\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y\pm {\frac {a^{2}}{c}}=0}$ in normal form.

An example using focus and directrix

Figure 2: Ellipse (red curve) at origin with major axis oblique.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Focus at point ${\displaystyle F}$${\displaystyle :(-20,15)}$
Directrix is line ${\displaystyle DE}$${\displaystyle :{\frac {4}{5}}x-{\frac {3}{5}}y+36=0.}$
eccentricity ${\displaystyle e={\frac {5}{6}}={\frac {P_{1}F}{P_{1}D}}={\frac {P_{2}F}{P_{2}E}}.}$
Ellipse has equation:
${\displaystyle 20x^{2}+24xy+27y^{2}-9900=0.}$

Given focus ${\displaystyle (-20,15),\ e={\frac {5}{6}}}$ and directrix with equation ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+36=0}$, calculate equation of the ellipse in form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$

See Figure 2. Let point ${\displaystyle P\ (x,y)}$ be any point on the ellipse.

Distance from ${\displaystyle P}$ to focus ${\displaystyle ={\sqrt {(x+20)^{2}+(y-15)^{2}}}}$

Distance from ${\displaystyle P}$ to directrix ${\displaystyle =0.8x-0.6y+36}$.

${\displaystyle {\sqrt {(x+20)^{2}+(y-15)^{2}}}={\frac {5}{6}}(0.8x-0.6y+36)}$.

${\displaystyle 6{\sqrt {(x+20)^{2}+(y-15)^{2}}}=5(0.8x-0.6y+36)=4x-3y+180}$.

${\displaystyle 36((x+20)^{2}+(y-15)^{2})=(4x-3y+180)^{2}}$.

${\displaystyle 36((x+20)^{2}+(y-15)^{2})-(4x-3y+180)^{2}=0}$.

Expand and the result is: ${\displaystyle 20x^{2}+24xy+27y^{2}+(0)x+(0)y-9900=0.}$

Because ${\displaystyle D=E=0,}$ the center of the ellipse is at the origin and the various lines have equations as follows.

Minor axis: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+0=0}$

Other directrix: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y-36=0}$

Major axis: ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+0=0}$

If you reverse-engineer the ellipse using the method above, ${\displaystyle K=25}$,

the expression ${\displaystyle KA=a^{2}-p^{2}}$ becomes ${\displaystyle (25)(20)=30^{2}-20^{2}}$, and

the expression ${\displaystyle KF=-a^{2}b^{2}}$ becomes ${\displaystyle (25)(-9900)=-(30^{2})(275)}$.

## General Ellipse

Figure 1: The general ellipse (red curve).
Foci at ${\displaystyle S_{1},S_{2},}$ center at ${\displaystyle C_{2}.}$
Point ${\displaystyle V}$ is a vertex. Length ${\displaystyle VC_{2}=a.}$
${\displaystyle PS_{1}+PS_{2}=2a.}$

The ellipse may assume any position and any orientation in Cartesian two-dimensional space. See the red curve in Figure 1.

The equation of this curve may be derived as follows:

Given two ${\displaystyle foci:\ S_{1}(s,t),\ S_{2}(u,v),}$ ${\displaystyle major\ axis}$ of length ${\displaystyle 2a}$ and point ${\displaystyle P(x,y)}$

${\displaystyle PS_{1}+PS_{2}=2a}$

${\displaystyle {\sqrt {(x-s)^{2}+(y-t)^{2}}}+{\sqrt {(x-u)^{2}+(y-v)^{2}}}=2a}$

The expansion of this expression is somewhat complicated because it contains 5 variables ${\displaystyle s,t,u,v,a.}$

The expansion may be simplified by reducing the number of variables from ${\displaystyle 5}$ to ${\displaystyle 3}$, the familiar ${\displaystyle p,q,a.}$

Let ${\displaystyle C_{2}}$, the center of the ellipse, have coordinates ${\displaystyle (G,H)}$ where ${\displaystyle G={\frac {s+u}{2}};\ H={\frac {t+v}{2}}}$ and ${\displaystyle p=u-G;\ q=v-H.}$

Then ${\displaystyle (s,t)=(G-p,H-q),}$ ${\displaystyle (u,v)=(G+p,H+q),}$ and

${\displaystyle {\sqrt {(x-(G-p))^{2}+(y-(H-q))^{2}}}+{\sqrt {(x-(G+p))^{2}+(y-(H+q))^{2}}}=2a}$

The expansion is: ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ where:

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

${\displaystyle B=-2pq}$

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

${\displaystyle D=2(Gp^{2}+Hpq-Ga^{2})}$

${\displaystyle E=2(Hq^{2}+Gpq-Ha^{2})}$

${\displaystyle F=(ap)^{2}+(aq)^{2}-a^{4}+(G^{2}+H^{2})a^{2}-(Gp+Hq)^{2}}$

A different approach

Begin with ellipse at origin: ${\displaystyle Ax^{2}+Bxy+Cy^{2}+F=0}$ where:

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

${\displaystyle B=-2pq}$

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

${\displaystyle F=(ap)^{2}+(aq)^{2}-a^{4}}$

By translation of coordinates, move the ellipse so that the new center of the ellipse is: ${\displaystyle (G,H).\ x}$${\displaystyle (x-G),\ y}$${\displaystyle (y-H).}$

The equation above becomes: ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F^{1}=0}$ where:

${\displaystyle D=-(2AG+BH);\ E=-(BG+2CH);\ F^{1}=AG^{2}+BGH+CH^{2}+F.}$

${\displaystyle D=-(2AG+BH)=-(2(a^{2}-p^{2})G+(-2pq)H)=-(2Ga^{2}-2Gp^{2}-2pqH)=2(Gp^{2}+Hpq-Ga^{2}),}$

the same as the value of ${\displaystyle D}$ in the method above.

The expansion of ${\displaystyle E,F^{1}}$ will show that this method produces the same results as the method above.

Given foci and major axis, perhaps the simplest way to produce ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F^{1}}$ is to calculate ${\displaystyle Ax^{2}+Bxy+Cy^{2}+F}$ and move the center from ${\displaystyle C_{1}:\ (0,0)}$ to ${\displaystyle C_{2}:\ (G,H)}$.

Center of ellipse

Given ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,}$ we know from above that ${\displaystyle D=-(2AG+BH);\ E=-(BG+2CH).}$

Therefore ${\displaystyle G={\frac {BE-2CD}{4AC-B^{2}}},\ H={\frac {-(E+BG)}{2C}}}$ where the point ${\displaystyle (G,H)}$ is the center of the ellipse.

## An Example of the General Ellipse

Figure 1: Translation of coordinate axes.
Red curve and green curve have same size, shape and orientation. The only difference is that center of green curve ${\displaystyle C_{1}(0,0)}$ has been moved to ${\displaystyle C_{2}(-34,14)}$ where it is the center of the red curve.
In both curves ${\displaystyle p=16;\ q=12.\ V_{1}C_{1}=V_{2}C_{2}=a=25.}$

See Figure 1. Given foci ${\displaystyle (-50,2),(-18,26)}$ and ${\displaystyle a=25,}$ calculate the equation of the ellipse in form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$

Calculate the center: ${\displaystyle G={\frac {-50+(-18)}{2}}=-34;\ H={\frac {2+26}{2}}=14.}$

${\displaystyle p=-18-(-34)=16;\ q=26-14=12.}$

Equation of ellipse at origin: ${\displaystyle 369x^{2}-384xy+481y^{2}-140625=0.}$

Move ellipse from origin ${\displaystyle C_{1}}$ to ${\displaystyle C_{2}\ (-34,14):}$

${\displaystyle 369x^{2}-384xy+481y^{2}+30468x-26524y+562999=0.}$

Reverse-engineering the general ellipse

Given ellipse ${\displaystyle 369x^{2}-384xy+481y^{2}+30468x-26524y+562999=0,}$ calculate the foci and the major axis.

Calculate the center of the ellipse (point ${\displaystyle C_{2}}$): ${\displaystyle (G,H)=(-34,14).}$

${\displaystyle KA-(a^{2}-p^{2})=0}$

${\displaystyle KB-(-2pq)=0}$

${\displaystyle KC-(a^{2}-q^{2})=0}$

${\displaystyle KF-((ap)^{2}+(aq)^{2}-a^{4}+(G^{2}+H^{2})a^{2}-(Gp+Hq)^{2})=0.}$

Solutions are:

${\displaystyle K={\frac {4F-4(AG^{2}+BGH+CH^{2})}{B^{2}-4AC}}}$

where ${\displaystyle A=369;\ B=-384;\ C=481;\ F=562999;\ G=-34;\ H=14.}$

In this example, ${\displaystyle K=1.}$

If ${\displaystyle (K}$ != ${\displaystyle 1)\ \{A}$${\displaystyle AK;\ B}$${\displaystyle BK;\ C}$${\displaystyle CK\}}$

The values ${\displaystyle p,q,a}$ may be calculated as in "Reverse-engineering ellipse at origin" above.

${\displaystyle p=16;\ q=12;\ a=25.}$

Length of major axis ${\displaystyle =2a=2(25)=50.}$

Focus ${\displaystyle S_{1}=(s,t)=(-34-16,14-12)=(-50,2).}$

Focus ${\displaystyle S_{2}=(u,v)=(-34+16,14+12)=(-18,26)}$

Significant lines of the Ellipse

Figure 2. Graph of ellipse illustrating axes, each latus rectum, each directrix.
Directrix through ${\displaystyle D_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+145=0}$
Directrix through ${\displaystyle D_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-105=0}$
Latus rectum through ${\displaystyle F_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+100=0}$
Latus rectum through ${\displaystyle F_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-60=0.}$
Major axis ${\displaystyle :\ y={\frac {4}{3}}x+100.}$
Minor axis ${\displaystyle :\ {\frac {3}{5}}x+{\frac {4}{5}}y+20=0.}$

The significant lines of the ellipse are: ${\displaystyle major\ axis,\ minor\ axis,}$ each ${\displaystyle latus\ rectum,}$ each ${\displaystyle directrix.}$

Consider the ellipse in Figure 2. Given foci ${\displaystyle F_{1}=(-108,-44),\ F_{2}=(-12,84)}$ and ${\displaystyle a=100,}$ calculate the equations of all the significant lines.

Slope of major axis ${\displaystyle ={\frac {84-(-44)}{-12-(-108)}}={\frac {128}{96}}={\frac {4}{3}}.}$

Major axis has equation ${\displaystyle y={\frac {4}{3}}x+g}$ and it passes through ${\displaystyle F_{1}.}$

Therefore, ${\displaystyle g=-44-{\frac {4}{3}}(-108)=-44+144=100.}$

Major axis ${\displaystyle V_{1}V_{2}}$ has equation: ${\displaystyle y={\frac {4}{3}}x+100.}$

Center of ellipse ${\displaystyle C=({\frac {-108+(-12)}{2}},{\frac {-44+84}{2}})=(-60,20).}$

Minor axis is perpendicular to major axis. Therefore, minor axis has equation: ${\displaystyle y=-{\frac {3}{4}}x+g}$ and it passes through the center ${\displaystyle C.}$

${\displaystyle g=20+{\frac {3}{4}}(-60)=20-45=-25.}$

Equation of minor axis (orange line through ${\displaystyle C}$): ${\displaystyle y=-{\frac {3}{4}}x-25}$ or ${\displaystyle 3x+4y+100=0}$ or ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20=0.}$

${\displaystyle F_{1}C=F_{2}C=c={\sqrt {(-60-(-108))^{2}+(20-(-44))^{2}}}={\sqrt {48^{2}+64^{2}}}=\pm 80.}$

Using the equation of the minor axis, the fact that each latus rectum is parallel to the minor axis, and that the distance from minor axis to latus rectum ${\displaystyle =c=80,}$ each latus rectum has equation: ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 80=0.}$

Equation of latus rectum (blue line) through ${\displaystyle F_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+100=0}$

Equation of latus rectum (blue line) through ${\displaystyle F_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-60=0.}$

Using the equation of the minor axis, the fact that each directrix is parallel to the minor axis, and that the distance from minor axis to directrix ${\displaystyle =D_{1}C=D_{2}C={\frac {a^{2}}{c}}={\frac {100^{2}}{80}}=125,}$ each directrix has equation: ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 125=0.}$

Equation of directrix (red line) through ${\displaystyle D_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+145=0}$

Equation of directrix (red line) through ${\displaystyle D_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-105=0.}$

## K and "Standard Form"

Figure 1: Three ellipses illustrating "standard form."

For green curve ${\displaystyle K=25.}$
For red curve ${\displaystyle K={\frac {1}{256}}.}$
For blue curve ${\displaystyle K=36.}$

Everything about the ellipse can be derived from ${\displaystyle G,H,\ p,q,a}$ the last three of which ${\displaystyle (p,q,a)}$ are contained within:

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

${\displaystyle B=-2pq}$

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

${\displaystyle F=-a^{2}b^{2}}$ for ellipse at origin.

Consider the ellipse: ${\displaystyle 369x^{2}-384xy+481y^{2}-3,515,625=0,}$ the green curve in Figure 1. It is tempting to say that ${\displaystyle p=16;\ q=12;\ a=25.}$

These values satisfy ${\displaystyle A=a^{2}-p^{2}=25^{2}-16^{2}=369;\ B=-2pq=-2(16)(12)=-384;\ C=a^{2}-q^{2}=25^{2}-12^{2}=481.}$

However, ${\displaystyle c^{2}=400;\ b^{2}=a^{2}-c^{2}=625-400=225;\ F=-a^{2}b^{2}=-(625)(225)=-140,625.}$ These values for ${\displaystyle p,q,a}$ are not correct.

Put the equation of the ellipse into "standard form." In this context "standard form" means that ${\displaystyle K=1.}$

For ellipse at origin ${\displaystyle K={\frac {4F}{B^{2}-4AC}}={\frac {4(-3,515,625)}{(-384)^{2}-4(369)(481)}}={\frac {-14,062,500}{-562,500}}=25.}$

In fact ${\displaystyle {\frac {3,515,625}{140,625}}=25=K.}$

${\displaystyle A}$${\displaystyle AK;\ B}$${\displaystyle BK;\ C}$${\displaystyle CK;\ F}$${\displaystyle FK.}$

${\displaystyle A=9,225;\ B=-9,600;\ C=12,025;\ F=-87,890,625.}$

The equation of the ellipse becomes: ${\displaystyle 9,225x^{2}-9,600xy+12,025y^{2}-87,890,625=0}$ and

${\displaystyle K={\frac {4F}{B^{2}-4AC}}={\frac {-351,562,500}{(-9600)^{2}-4(9225)(12025)}}={\frac {351562500}{351562500}}=1.}$

The equation of the ellipse is in "standard form" and:

${\displaystyle p=80;\ q=60;\ a=125.}$

${\displaystyle A=a^{2}-p^{2}=15,625-6,400=9,225.}$

${\displaystyle B=-2pq=-2(80)(60)=-9,600.}$

${\displaystyle C=a^{2}-q^{2}=15,625-3,600=12,025.}$

${\displaystyle c^{2}=p^{2}+q^{2}=80^{2}+60^{2}=100^{2};\ b^{2}=a^{2}-c^{2}=15,625-10,000=5,625;}$

${\displaystyle F=-a^{2}b^{2}=-15,625(5,625)=-87,890,625}$

The values ${\displaystyle p=80;\ q=60;\ a=125}$ are correct.

Example 2. Consider the ellipse ${\displaystyle 5,904x^{2}-6,144xy+7,696y^{2}-140,625=0,}$ the red curve in Figure 1.

In this example, ${\displaystyle K={\frac {1}{256}}=0.003,906,25}$ and the equation in "standard form" is:

${\displaystyle 23.0625x^{2}-24xy+30.0625y^{2}-549.316,406,25=0}$.

Example 3. Consider the ellipse ${\displaystyle 9x^{2}+25y^{2}-900x-2000y+54400=0,}$ the blue curve in Figure 1.

The center ${\displaystyle (G,H)=(50,40).}$

In this example ${\displaystyle B=0;\ K={\frac {AG^{2}+CH^{2}-F}{AC}}={\frac {9(50)^{2}+25(40^{2})-54400}{9(25)}}=36}$ and the equation in "standard form" is:

${\displaystyle 324x^{2}+900y^{2}-32400x-72000y+1958400=0}$.

## Tangent at latus rectum

Figure 1: Ellipse and tangent at Latus Rectum.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Red curve is ellipse at origin with major axis vertical.
Line ${\displaystyle D_{1}DD_{2}}$ is directrix: ${\displaystyle y=-{\frac {a^{2}}{c}}.}$
Line ${\displaystyle DR}$ passes through point ${\displaystyle (0,-{\frac {a^{2}}{c}}).}$
Line ${\displaystyle DR}$ is tangent to curve at Latus Rectum: ${\displaystyle ({\frac {b^{2}}{a}},-c).}$

See Figure 1. The red curve is that of an ellipse at the origin with major axis vertical: ${\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0.}$

The line ${\displaystyle D_{1}DD_{2}}$ is the directrix with equation: ${\displaystyle y=-{\frac {a^{2}}{c}}.}$

The green line ${\displaystyle DR}$ has equation: ${\displaystyle y=mx-{\frac {a^{2}}{c}}.}$

The aim of this section is to show that the line ${\displaystyle DR}$ is tangent to the ellipse at the ${\displaystyle latus\ rectum.}$

Let the line intersect the curve. The ${\displaystyle x}$ coordinates of the point of intersection are given by:

${\displaystyle (+aacc+bbccmm)xx+(-2aabbcm)x+(+aaaabb-aabbcc)=0}$

If the line ${\displaystyle DR}$ is a tangent, ${\displaystyle x}$ has one value and the discriminant is ${\displaystyle 0:}$

${\displaystyle (-2aabbcm)(-2aabbcm)-4(+aacc+bbccmm)(+aaaabb-aabbcc)=0}$ or:

${\displaystyle (+bbcc)mm+(-aaaa+aacc)=0.}$

${\displaystyle mm={\frac {aaaa-aacc}{bbcc}}={\frac {aa(aa-cc)}{bbcc}}={\frac {aabb}{bbcc}}={\frac {aa}{cc}}}$

${\displaystyle m={\sqrt {\frac {aa}{cc}}}=\pm {\frac {a}{c}}}$

The tangent ${\displaystyle DR}$ has slope ${\displaystyle {\frac {a}{c}}}$and equation: ${\displaystyle y={\frac {a}{c}}x-{\frac {a^{2}}{c}}.}$

Let this line intersect the curve. The ${\displaystyle x}$ coordinates of the points of intersection are given by: ${\displaystyle (+bb+cc)xx+(-2abb)x+(+aabb-bbcc)=0.}$

Discriminant = ${\displaystyle (-2abb)(-2abb)-4(+bb+cc)(+aabb-bbcc)=bb+cc-aa=0.}$

${\displaystyle x={\frac {-(-2abb)}{2(bb+cc)}}={\frac {abb}{aa}}={\frac {b^{2}}{a}}=}$ half length of latus rectum.

The tangent ${\displaystyle DR}$ touches the curve where ${\displaystyle x={\frac {b^{2}}{a}}}$, the point ${\displaystyle R}$ where ${\displaystyle chord\ LR}$ is the latus rectum.

## Reflectivity of ellipse

Figure 1: Ellipse (red curve) with major axis horizontal.

Origin at point ${\displaystyle O}$${\displaystyle :(0,0)}$.
Foci are points ${\displaystyle F_{1}(-c,0),\ F_{2}(c,0).}$
Line ${\displaystyle T_{1}PT_{2}}$ tangent to curve at ${\displaystyle P}$.
Angle of incidence = angle of reflection: ${\displaystyle \angle {F_{2}PT_{2}}=\angle {F_{1}PT_{1}}.}$

See Figure 1.

The curve (red line) is an ellipse with equation: ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$ where ${\displaystyle A=b^{2},\ C=a^{2}.}$

{\displaystyle {\begin{aligned}a^{2}y^{2}=a^{2}b^{2}-b^{2}x^{2}\\\\y^{2}={\frac {a^{2}b^{2}-b^{2}x^{2}}{a^{2}}}\\\\y={\sqrt {\frac {a^{2}b^{2}-b^{2}x^{2}}{a^{2}}}}={\frac {\sqrt {a^{2}b^{2}-b^{2}x^{2}}}{a}}\end{aligned}}}

Foci ${\displaystyle F_{1},F_{2}}$ have coordinates ${\displaystyle (-c,0),(c,0).}$

Line ${\displaystyle T_{1}PT_{2}}$ is tangent to the curve at point ${\displaystyle P.}$

A ray of light emanating from focus ${\displaystyle F_{2}}$ is reflected from the inside surface of the ellipse at point ${\displaystyle P}$ and passes through the other focus ${\displaystyle F_{1}.}$

The aim is to prove that ${\displaystyle \angle {F_{1}PT_{1}}=\angle {F_{2}PT_{2}}.}$

Point ${\displaystyle N}$ has coordinates ${\displaystyle (c+u,0).}$

At point ${\displaystyle P,\ x=c+u,\ y={\frac {\sqrt {a^{2}b^{2}-b^{2}x^{2}}}{a}}={\frac {\sqrt {a^{2}b^{2}-b^{2}(c+u)^{2}}}{a}}={\frac {R}{a}}}$

Slope of line ${\displaystyle F_{2}P={\frac {PN}{F_{2}N}}={\frac {R}{au}}=m_{2}.}$

Slope of line ${\displaystyle F_{1}P={\frac {PN}{F_{1}N}}={\frac {R}{a(2c+u)}}=m_{1}.}$

Slope of curve at ${\displaystyle P={\frac {-Ax}{Cy}}={\frac {-A(c+u)}{C(R/a)}}={\frac {-A(c+u)a}{CR}}=m.}$

Using ${\displaystyle \tan(A-B)={\frac {\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}}}$,

${\displaystyle \tan(\angle {T_{1}PF_{1}})={\frac {m_{1}-m}{1+m_{1}m}}={\frac {aA(C+cc+cu)}{R(C(2c+u)-A(c+u))}}}$

${\displaystyle \tan(\angle {T_{2}PF_{2}})={\frac {m-m_{2}}{1+mm_{2}}}={\frac {aA(-C+cc+cu)}{R(Cu-A(c+u))}}}$

if ${\displaystyle \angle {T_{1}PF_{1}}==\angle {T_{2}PF_{2}}}$ then:

${\displaystyle \tan(\angle {T_{1}PF_{1}})=\tan(\angle {T_{2}PF_{2}})}$,

${\displaystyle {\frac {aA(C+cc+cu)}{R(C(2c+u)-A(c+u))}}={\frac {aA(-C+cc+cu)}{R(Cu-A(c+u))}}}$,

${\displaystyle (C+cc+cu)(Cu-A(c+u))=(C(2c+u)-A(c+u))(-C+cc+cu)}$ and

${\displaystyle (C+cc+cu)(Cu-A(c+u))-(C(2c+u)-A(c+u))(-C+cc+cu)=0}$ where ${\displaystyle A=bb,\ C=aa,\ cc=aa-bb.}$

If you make the substitutions and expand, the result is ${\displaystyle 0}$.

Therefore, angle of reflection ${\displaystyle \angle {F_{1}PT_{1}}=}$ angle of incidence ${\displaystyle \angle {F_{2}PT_{2}}}$ and the reflected ray ${\displaystyle PF_{1}}$ passes through the other focus ${\displaystyle F_{1}.}$