A tangent is drawn to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  to cut  $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$  at points P and Q. If tangent at P and Q to the ellipse  $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$  intersect at right angle then  $\frac{a^2}{c^2}+\frac{b^2}{d^2}=$ .

Let tangents to  $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$  at P and Q intersect at (h, k) at right angle
$\therefore h^2+k^2=c^2+d^2$  (director circle) ----------------(1)
And chord of contact PQ is given by
$\frac{hx}{c^2}+\frac{ky}{d^2}=1$ 
 Which is tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is--------------(2)
Equation of  tangent to  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$         --------------------------(3)
Equation (2) and (3) represent the same line
$\therefore \frac{ha}{c^2\cos \theta }+\frac{kb}{d^2\sin \theta }=1$ 
 $\therefore \frac{h^2{a}^2}{c^4}+\frac{k^2{b}^2}{d^4}=1$
Comparing with equation (1), 
 $\frac{a^2}{c^4}=\frac{b^2}{d^4}=\frac{1}{c^2+d^2}\Rightarrow \frac{a^2}{c^2}=\frac{c^2}{c^2+d^2}\&\frac{b^2}{d^2}=\frac{1}{c^2+d^2}$
Hence  $\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{c^2+d^2}{c^2+d^2}=1$