Question

Show that the locus of the intersection of two perpendicular normals to an ellipse is the curve
$(a^2+b^2)(x^2+y^2){(a^2{y}^2+b^2{x}^2)}^2={(a^2-b^2)}^2{(a^2{y}^2-b^2{x}^2)}^2$.

Answer 1

As the normals are perpendicular, therefore the corresponding tangents are also perpendicular. So they must intersect on the director circle which is given by $x^2+y^2=a^2+b^2=A^2$ (say)
Any point P on the director circle may be taken as, $(A\cos \theta ,A\sin \theta )$
Let the equation to the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
So the polar of P w.r.t. the ellipse will be $\frac{Ax\cos \theta }{a^2}+\frac{Ay\sin \theta }{b^2}=1$
The intersection of the perpendicular normals may be given by the equation
So putting $l=\frac{A\cos \theta }{a^2}$ and $m=\frac{A\sin \theta }{b^2}$ we get,
$x=(a^2-b^2)\frac{A\cos \theta }{a^2}\frac{1-\frac{(a^2+b^2)}{b^2}{\sin }^2\theta }{\frac{a^2+b^2}{a^2}-{\cos }^2\theta }+\left\{\frac{(a^2+b^2){\sin }^2\theta }{b^2}\right\}$
$\therefore x=\frac{a^2-b^2}{a^2+b^2}\frac{b^2{\cos }^2\theta -a^2{\sin }^2\theta }{b^2{\cos }^2\theta +a^2{\sin }^2\theta }A\cos \theta ....1$
and $y=-(a^2-b^2)\frac{A\sin \theta }{b^2}\frac{1-\left(\frac{a^2+b^2}{a^2}\right){\cos }^2\theta }{\frac{(a^2+b^2)}{a^2}{\cos }^2\theta +\frac{(a^2+b^2)}{b^2}{\sin }^2\theta }$
$=\frac{a^2-b^2}{a^2+b^2}\frac{b^2{\cos }^2\theta -a^2{\sin }^2\theta }{b^2{\cos }^2\theta +a^2{\sin }^2\theta }A\sin \theta .....2$
$\frac{x}{y}=\frac{\cos \theta }{\sin \theta }$
$\frac{\sin \theta }{y}=\frac{\cos \theta }{x}=\frac{1}{\sqrt{x^2+y^2}}.....3$
Eliminating from $\theta$ from $1$ and $3$, we get$x=\frac{a^2-b^2}{a^2+b^2}A\frac{x}{\sqrt{x^2+y^2}}\frac{b^2{x}^2-a^2{y}^2}{b^2{x}^2+a^2{y}^2}$
or ${(a^2+b^2)}^2(x^2+y^2){(b^2{x}^2+a^2{y}^2)}^2={(a^2-b^2)}^2{(b^2{x}^2-a^2{y}^2)}^2$