If the eccentric angles of two points P and Q on the  ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ are $\alpha ,\beta$ such that $\alpha +\beta =\frac{\pi }{2}$ then,the locus of the point of intersection of the normals at  $P$ and $Q$ is 

Equations of the normal at$P$  and$Q$  are

$ax \sec \alpha -$by $\mathrm{cosec}\alpha =a^2-b^2$

$ax\sec \beta -$ by $\mathrm{cosec}\beta =a^2-b^2$

If they intersect at $(h, k).$

$ah\sec \alpha -bk\mathrm{cosec}\alpha =ah\mathrm{cosec}\alpha -bk\sec \alpha$

$(\because \beta =\pi /2-\alpha )$ 

$(ah+bk)(\sec \alpha -\mathrm{cosec}\alpha )=0\Rightarrow ah+bk=0$

Locus of  $(h,k)$ is $ax+by=0$