Ellipse

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Ellipse

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

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

In cartesian geometry in two dimensions the ellipse is the locus of a point $P$ that moves relative to two fixed points called $foci$$:F_{1},F_{2}.$ The distance $F_{1}F_{2}$ from one $focus\ (F_{1})$ to the other $focus\ (F_{2})$ is non-zero. The sum of the distances $(PF_{1},PF_{2})$ from point to foci is constant.

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

The center of the ellipse is located at the origin $O(0,0)$ and the foci $(F_{1},F_{2})$ are on the $X\ axis$ at distance $c$ from $O.$

$F_{1}$ has coordinates $(-c,0).F_{2}$ has coordinates $(c,0)$. Line segments $OF_{1}=OF_{2}=c.$

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

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

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

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

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

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

Equation of the ellipse

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

Let point $P$ have coordinates $(x,y).$

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

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

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

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

$M+N=2a$

$MM+2MN+NN=4aa$

$2MN=4aa-MM-NN$

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

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

Make appropriate substitutions, expand and the result is:

$16aaxx-16ccxx+16aayy-16aaaa+16aacc=0$

$(aa-cc)xx+aayy-aa(aa-cc)=0$

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

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

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

Length of latus rectum

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

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

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

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

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

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

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

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

The directrices

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

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

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

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 $e=eccentricity,\ 0<e<1.$

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

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

Consider the point $V_{1}$:

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

$2ae+a-c=a+c;\ 2ae=2c.$

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

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

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

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

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

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

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

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

Ellipse using focus and directrix

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

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

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

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

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

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

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

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

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

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

Expand and the result is:

$+eexx+eeyy-eeeeyy+ccee-cc=0$

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

$xx+(1-ee)yy+cc-aa=0$

$xx+(1-ee)yy-(aa-cc)=0$

$xx+(1-ee)yy-bb=0$

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

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

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

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

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

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

General ellipse at origin

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

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

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

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

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

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

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

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

$M+N=2a$

$MM+2MN+NN=4aa$

$2MN=4aa-MM-NN$

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

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

Make appropriate substitutions, expand and the result is:

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

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

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

$B=-2pq$

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

$D=E=0$ because center is at origin,

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

An example

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Let $(p,q)$ be (3,4) and $a=8.$

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

Reverse-engineering ellipse at origin

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

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

$B=-2pq$

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

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

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

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

To produce consistent, correct values for $p,q,a$ the equations become:

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

$KB=-2pq$

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

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

or:

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

$KB-(-2pq)=0$

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

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

Solutions are:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

An example using focus and directrix

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

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

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

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

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

Distance from $P$ to directrix $=0.8x-0.6y+36$.

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

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

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

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

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

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

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

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

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

If you reverse-engineer the ellipse using the method above, $K=25$,

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

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

General Ellipse

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

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

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

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

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

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

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

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

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

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

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

$B=-2pq$

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

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

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

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

A different approach

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

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

$B=-2pq$

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

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

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

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

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

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

the same as the value of $D$ in the method above.

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

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

Center of ellipse

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

Therefore $G={\frac {BE-2CD}{4AC-B^{2}}},\ H={\frac {-(E+BG)}{2C}}$ where the point $(G,H)$ 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 $C_{1}(0,0)$ has been moved to $C_{2}(-34,14)$ where it is the center of the red curve.
In both curves $p=16;\ q=12.\ V_{1}C_{1}=V_{2}C_{2}=a=25.$

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

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

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

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

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

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

Reverse-engineering the general ellipse

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

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

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

$KB-(-2pq)=0$

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

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

Solutions are:

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

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

In this example, $K=1.$

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

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

$p=16;\ q=12;\ a=25.$

Length of major axis $=2a=2(25)=50.$

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

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

Significant lines of the Ellipse

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

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

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

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

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

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

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

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

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

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

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

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

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 $=c=80,$ each latus rectum has equation: ${\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 80=0.$

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

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

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 $=D_{1}C=D_{2}C={\frac {a^{2}}{c}}={\frac {100^{2}}{80}}=125,$ each directrix has equation: ${\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 125=0.$

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

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

K and "Standard Form"

Figure 1: Three ellipses illustrating "standard form."

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

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

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

$B=-2pq$

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

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

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

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

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

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

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

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

$A$ ← $AK;\ B$ ← $BK;\ C$ ← $CK;\ F$ ← $FK.$

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

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

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

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

$p=80;\ q=60;\ a=125.$

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

$B=-2pq=-2(80)(60)=-9,600.$

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

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

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

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

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

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

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

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

The center $(G,H)=(50,40).$

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

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

Tangent at latus rectum

Figure 1: Ellipse and tangent at Latus Rectum.

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

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

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

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

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

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

$(+aacc+bbccmm)xx+(-2aabbcm)x+(+aaaabb-aabbcc)=0$

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

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

$(+bbcc)mm+(-aaaa+aacc)=0.$

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

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

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

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

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

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

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

Reflectivity of ellipse

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

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

See Figure 1.

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

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

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

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

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

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

Point $N$ has coordinates $(c+u,0).$

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

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

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

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

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

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

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

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

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

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

$(C+cc+cu)(Cu-A(c+u))=(C(2c+u)-A(c+u))(-C+cc+cu)$ and

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

If you make the substitutions and expand, the result is $0$.

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