Ellipse

Ellipse
[[File:20170520 Ellipse Horizontal At Origin 02.png|thumb|400px| 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 $$V_1OV_2$$ is the $$major\ axis.$$ $$\ \ \ \ \ OV_1 = OV_2 = B_1F_1 = B_1F_2 = a.$$ Line segment $$B_1OB_2$$ is the $$minor\ axis.\ OB_1 = OB_2 = b.$$ Each line $$L_1F_1R_1, L_2F_2R_2$$ 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 $$PF_1+PF_2 = B_1F_1+B_1F_2 = V_1F_1+V_1F_2 = K.$$

$$V_1F_1 = V_2F_2.\ \therefore V_1F_1+V_1F_2 = V_1F_2 + V_2F_2 = V_1V_2 = K =2a,$$ the length of the $$major\ axis\ (V_1V_2).\ OV_1 = OV_2 = a.$$

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 $$B_1B_2$$ perpendicular to the major axis at the midpoint of the major axis is the $$minor\ axis$$ with length $$2b.\ OB_1 = OB_2 = b.$$

$$B_1F_1 + B_1F_2 = 2a;\ B_1F_1 = B_1F_2 = a.\ \therefore a^2 = b^2 + c^2.$$

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 $$L_1F_1R_1,\ L_2F_2R_2$$ perpendicular to the major axis is a $$latus\ rectum.$$

Equation of the 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:

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

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

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

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

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

Length of latus rectum ---

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

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

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

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

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

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

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

Length of latus rectum $$= L_1R_1 = L_2R_2 = \frac{2b^2}{a}.$$

The directrices
[[File:20170519 Ellipse Illustrating Eccentricity 03.png|thumb|400px| 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 $$x = D_1,\ x = D_2,\ D_1 = -D_2.$$ $$OV_1 = OV_2 = a.$$ $$eccentricity\ e = \frac{c}{a} = \frac{F_2V_2}{V_2D_2} = \frac{V_1F_2}{V_1D_2} = \frac{OV_2}{OD_2}$$ $$e = \frac{PF_1}{PE_1}$$ (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. $$\frac{F_2V_2}{V_2D_2} = e.$$

Let the length $$V_2D_2 = d.\ \frac{a-c}{d} = e.\ d= \frac{a-c}{e}$$

Consider the point $$V_1$$:

$$\frac{V_1F_2}{V_1D_2} = \frac{a+c}{2a+d} = e = (a+c)/(2a+\frac{a-c}{e}) = (a+c)/(\frac{2ae+a-c}{e}) = \frac{e(a+c)}{2ae+a-c}.$$

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

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

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 = $$OF_2 + F_2D_2 = c + d_0 = \frac{c}{e^2}$$.

$$e^2c + e^2d_0 = c;\ c - e^2c = e^2d_0;\ c = \frac{e^2d_0}{1-e^2}$$.

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

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

Distance from focus to directrix $$= F_2D_2 = \frac{b^2}{c}$$.

Ellipse using focus and directrix

[[File:20170520 EllipseAtOriginVertical01.png|thumb|300px| 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:

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

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

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

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

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

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

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

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

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

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

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

General ellipse at origin
[[File:20170516 General Ellipse At Origin 03.png|thumb|300px| 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 $$V_1OV_2$$ is the $$major\ axis.$$ $$\ \ \ \ \ OV_1 = OV_2 = a.$$ Line $$D_1OD_2$$ is the major axis extended. Lines perpendicular to $$D_1OD_2$$ at $$D_1, F_1, O ,F_2, D_2$$ are directrix, latus rectum. minor axis, latus rectum, directrix. $$OD_1 = OD_2 = \frac{a^2}{c}.$$ $$F_1D_1 = F_2D_2 = \frac{b^2}{c}.$$ $$PF_1 + PF_2 = 2a.$$ ]]

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

The line $$V_1V_2$$ 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,

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

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.$$)$$

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

$$B = -2pq$$

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

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

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:

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

$$KB = -2pq$$

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

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

or:

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

$$KB - (-2pq) = 0$$

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

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

Solutions are:

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

$$4(pp)^2 + 4K(A-C)pp - B^2K^2 = 0$$ 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 -

[[File:20170523 Ellipse at origin 00.png|thumb|300px| 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 $$e = \frac{5}{6} = \frac{P_1F}{P_1D} = \frac{P_2F}{P_2E}.$$ 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 $$KF = -a^2b^2$$ 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
[[File:20170531 General ellipse 03.png|thumb|350px| 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 $$p = 16;\ q =12.\ V_1C_1 = V_2C_2 = a = 25.$$

]]

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

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 $$V_1V_2$$ 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 $$\frac{3}{5}x + \frac{4}{5}y + 20 = 0.$$

$$F_1C = F_2C = c = \sqrt{(-60-(-108))^2 + (20-(-44))^2} = \sqrt{48^2 + 64^2} = \pm 80.$$

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 $$= D_1C = D_2C = \frac{a^2}{c} = \frac{100^2}{80} = 125,$$ 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"
[[File:20170602 Ellipses illustrating K 01.png|thumb|400px| 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:

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

$$B = -2pq$$

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

$$F = -a^2b^2$$ 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, $$c^2 = 400;\ b^2 = a^2-c^2 = 625-400=225;\ F=-a^2b^2 = -(625)(225) = -140,625.$$ 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:

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

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

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

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

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

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

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
[[File:20170603 Ellipse & tangent at LR 00.png|thumb|400px| 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 $$D_1DD_2$$ 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: $$a^2x^2 + b^2y^2 - a^2b^2 = 0.$$

The line $$D_1DD_2$$ 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
[[File:20170516 Reflectivity Of Ellipse 02.png|thumb|500px| 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 $$T_1PT_2$$ tangent to curve at $$P$$. Angle of incidence = angle of reflection: $$\angle{F_2PT_2} = \angle{F_1PT_1}.$$ ]]

See Figure 1.

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


 * $$\begin{align}

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

\end{align}$$

Foci $$F_1, F_2$$ have coordinates $$(-c,0), (c,0).$$

Line $$T_1PT_2$$ 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 $$\angle{F_1PT_1} = \angle{F_2PT_2}.$$

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

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

Slope of line $$F_2P = \frac{PN}{F_2N} = \frac{R}{au} = m_2.$$

Slope of line $$F_1P = \frac{PN}{F_1N} = \frac{R}{a(2c+u)} = m_1.$$

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

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

$$\tan (\angle{T_1PF_1}) = \frac{m_1 - m}{1+m_1m} = \frac { aA(C + cc + cu)}  { R(C(2c + u) - A(c + u))}       $$

$$\tan (\angle{T_2PF_2}) = \frac{m - m_2}{1+mm_2} = \frac {aA(-C + cc + cu) }  { R(Cu - A(c + u)) }       $$

if $$\angle{T_1PF_1} == \angle{T_2PF_2}$$ then:

$$\tan(\angle{T_1PF_1}) = \tan(\angle{T_2PF_2})$$,

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

$$  (C + cc + cu)   (Cu - A(c + u)) =  (C(2c + u) - A(c + u)) (-C + cc + cu) $$ 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 $$\angle{F_1PT_1} = $$ angle of incidence $$\angle{F_2PT_2}$$ and the reflected ray $$PF_1$$ passes through the other focus $$F_1.$$