Ellipse "inversion"

The locus of the point $F$ is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ .

Proof :- Let at any position of $A$ , it's coordinates be $(h, 0)$ . Then those of $D$ are $(h, \dfrac{b}{a}\sqrt {a^2-h^2})$ and of $E$ are $(h, -\dfrac{b}{a}\sqrt {a^2-h^2})$ . Therefore the equation of $\overline {DB}$ is $y=-\dfrac{b}{a}\sqrt {\dfrac{a+h}{a-h}}(x-a)$ , and of $\overline {CE}$ is $y=-\dfrac{b}{a}\sqrt {\dfrac{a-h}{a+h}}(x+a)$ . Hence the coordinates of $F$ are $(\dfrac{a^2}{h}, -\dfrac{b}{h}\sqrt {a^2-h^2})$ . Therefore the locus of $F$ is $y^2=\dfrac{b^2x^2}{a^2}-b^2$ or $\boxed {\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}$

Since $B$ , $C$ , $D$ and $E$ are on the ellipse, we can assign their coordinates as $B(a,0)$ , $C(-a,0)$ , $D (a\cos \theta, b \sin \theta)$ and $E(a\cos \theta, - b \sin \theta)$ . Then the equations for lines $DF$ and $CF$ are:

$\begin{cases} DF: & \dfrac {y-0}{x-a} = \dfrac {b\sin \theta - 0}{a\cos \theta - a} & \implies y = \dfrac {b\sin \theta(x-a)}{a(\cos \theta -1)} \\ CF: & \dfrac {y-0}{x+a} = \dfrac {-b\sin \theta - 0}{a\cos \theta + a} & \implies y = \dfrac {b\sin \theta(x+a)}{a(\cos \theta +1)} \end{cases}$

Then the coordinates of $F$ are given by:

$\begin{aligned} \frac {b\sin \theta(x_F-a)}{a(\cos \theta -1)} & = \frac {-b\sin \theta(x_F+a)}{a(\cos \theta +1)} \\ (a-x_F)(\cos \theta +1) & = (a+x_F)(\cos \theta -1) \\ \implies x_F & = \frac a{\cos \theta} = a \sec \theta \\ DF: \qquad \qquad y_F & = \frac {b\sin \theta (a\sec \theta -a)}{a(\cos \theta -1)} \\ & = \frac {b\sin \theta(1-\cos \theta)}{\cos \theta (\cos \theta -1)} \\ & = - b \tan \theta \end{aligned}$

Then $\dfrac {x_F^2}{a^2} - \dfrac {y_F^2}{b^2} = \sec^2 \theta - \tan^2 \theta = 1$ , which is a hyperbola .