# Ellipse
## Ellipse

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 ellipse

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

$b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0$
$b^{2}c^{2}+a^{2}y^{2}-a^{2}b^{2}=0$
$b^{2}(a^{2}-b^{2})+a^{2}y^{2}-a^{2}b^{2}=0$
$b^{2}a^{2}-b^{4}+a^{2}y^{2}-a^{2}b^{2}=0$
$a^{2}y^{2}=b^{4}$
$y^{2}={\frac {b^{4}}{a^{2}}}$
$y={\frac {b^{2}}{a}}$
Length of latus rectum $=L_{1}R_{1}=L_{2}R_{2}={\frac {2b^{2}}{a}}.$

## Second definition of ellipse

* Graph of ellipse ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$ where $a,b=20,12$.
* At point $B,\ {\frac {u}{v}}=e.$
* At point $A,\ {\frac {a}{t}}=e.$
* Ellipse is path of point that moves so that ratio of distance to
* fixed point and distance to fixed line is constant.
* Let ${\frac {p}{a}}=e$ where:
* $p$ is non-zero,
* $a>p,$
* $a=p+u.$
* Therefore, $1>e>0.$
* Let directrix have equation $x=t$ where ${\frac {a}{t}}=e.$
* At point $B:$
* ${\frac {p}{p+u}}={\frac {p+u}{p+u+v}}=e$
* $(p+u)^{2}=p(p+u+v)$
* $pp+pu+pu+uu=pp+pu+pv$
* $pu+uu=pv$
* $u(p+u)=pv$
* ${\frac {u}{v}}={\frac {p}{p+u}}=e$
* ${\frac {\text{distance to focus}}{\text{distance to directrix}}}=e\ \dots \ (3)$
* Statement $(3)$ is true at point $A$ also.
* Section under "Proof" below proves that statement (3) is true for any point $P$ on ellipse.

### Proof

* Proving that ${\frac {\text{distance from point to focus}}{\text{distance from point to directrix}}}=e$.
* Graph is part of curve ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$
* distance to Directrix1 $=t-x={\frac {a}{e}}-x={\frac {a-ex}{e}}.$
* base = $x-p=x-ae$
* ${\text{(distance to Focus1)}}^{2}={\text{base}}^{2}+y^{2}$
* As expressed above in statement $3,$ second definition of ellipse states that ellipse 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 $P$ on the ellipse.
* At point $P:$
* $(a^{2}-p^{2})x^{2}+a^{2}y^{2}-a^{2}(a^{2}-p^{2})=0$
* $y^{2}={\frac {-(a^{2}-p^{2})x^{2}+a^{2}(a^{2}-p^{2})}{a^{2}}}$
* $={\frac {a^{2}e^{2}x^{2}-a^{2}x^{2}+a^{2}a^{2}-a^{2}a^{2}e^{2}}{a^{2}}}$
* $=e^{2}x^{2}-x^{2}+a^{2}-a^{2}e^{2}$
* base $=x-p=x-ae$
* $({\text{distance}}\ F_{1}P)^{2}=y^{2}+{\text{base}}^{2}=y^{2}+(x-ae)^{2}$
* $=a^{2}-2aex+e^{2}x^{2}$
* $=(a-ex)^{2}$
* ${\text{distance to Focus1}}={\text{distance}}\ F_{1}P=a-ex$
* ${\text{distance to Directrix1}}=t-x={\frac {a}{e}}-x={\frac {a-ex}{e}}$
* ${\frac {\text{distance to Focus1}}{\text{distance to Directrix1}}}$
* $=(a-ex){\frac {e}{(a-ex)}}$
* $=e$
* Similar calculations can be used to prove the case for Focus2 $(-p,0)$ and Directrix2 $(x=-t)$
* in which case:
* ${\frac {\text{distance to Focus2}}{\text{distance to Directrix2}}}$
* $=(a+ex){\frac {e}{(a+ex)}}$
* $=e$
* Therefore: ${\frac {\text{distance to focus}}{\text{distance to directrix}}}=e$ where $1>e>0.$
* Ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant,
* called eccentricity $e.$

## The directrices

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

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

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
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
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
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

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
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
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

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
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
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"

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

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

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 $F_{1}.$

## Links to related Topics

* Conic sections
* Quartic function#Two Conic Sections