Question

Prove that the loci of the point of intersection of normals at the ends of focal chords of an ellipse are the two ellipses
$a^2{y}^2{(1+e^2)}^2+b^2(x\pm ae)(x\mp a{e}^3)=0$.

Answer 1

Let the equation to the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and any chord be $lx+my=1$. If the chord passes through the focus of the ellipse, i.e., $(ae,0)$,
We have $ael=1$
Now, let $(h.k)$ be the co ordinates of the points of intersection of normals; then by Art $412$, we get
$h=l(a^2-b^2)\frac{1-b^2{m}^2}{a^2{l}^2+b^2{m}^2}=\frac{l{a}^2{e}^2(1-b^2{m}^2)}{\frac{1}{e^2}+b^2{m}^2}$
where, $\frac{h}{ae}=\frac{e^2-e^2{b}^2{m}^2}{1+e^2{b}^2{m}^2}\because ael=1$
Similarly
$\frac{k}{m{a}^2{m}^2}=\frac{1-e^2}{1+e^2{b}^2{m}^2}$
or $k=\frac{a^2(1-e^2)e^{2}m}{1+e^2{b}^2{m}^2}=\frac{b^2{e}^2{m}^2}{1+e^2{b}^2{m}^2}$
So, $\frac{h+ae}{ae}=\frac{1+e^2}{1+e^2{b}^2{m}^2}$
and $\frac{h-a{e}^2}{ae}=-\frac{e^2{b}^2{m}^2(1+e^2)}{1+e^2{b}^2{m}^2}$
$\frac{(h+ae)(h-a{e}^3)}{a^1{e}^2}=-\frac{{(1+e^2)}^2(+e^2{b}^2{m}^2)}{{(1+e^2{b}^2{m}^2)}^2}$
$\frac{(h+ae)(h-a{e}^3)}{a^2{e}^2}=-\frac{{(1+e^2)}^2{a}^2{k}^2}{a^2{b}^2{e}^2}$
$a^2{k}^2{(1+e^2)}^2+b^2(h+ae)(h-a{e}^3)=0$
Locus of $(h,k)$ is;-
$a^2{y}^2{(1+e^2)}^2+b^2(x+ae)(x-a{e}^3)=0$
Similarly, locus of the chord can be obtained for $(-ae,0)$