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 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 $=L_{1}R_{1}=L_{2}R_{2}={\frac {2b^{2}}{a}}.$

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

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 $F_{1}.$

Links to related Topics

• Conic sections
• Quartic function#Two Conic Sections