If $sin^{-1}\frac{x}{a}+sin^{-1}\frac{y}{b}= \alpha $, then 

$\frac{x^2}{a^2}+\frac{2xy}{ab} cos \alpha +\frac{y^2}{b^2}= sin^2 \alpha $

We have, 

$sin^{-1}\frac{x}{a}+sin^{-1}\frac{y}{b}= \alpha $

$⇒ \frac{\pi}{2}- cos^{-1}\frac{x}{a}+\frac{\pi}{2}-cos^{-1}\frac{y}{b}=\alpha $

$⇒ cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \pi - \alpha $

$⇒ cos^{-1}\begin{Bmatrix}\frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}\end{Bmatrix} = \pi - \alpha $

$⇒ \frac{xy}{ab}- \sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}}= cos (\pi - \alpha )$

$⇒ \left(\frac{xy}{ab}+cos \alpha \right)^2= \left(1-\frac{x^2}{a^2}\right) \left(1-\frac{y^2}{b^2}\right)$

$⇒\frac{x^2}{a^2}+\frac{2xy}{ab}cos \alpha +\frac{y^2}{b^2}= sin^2 \alpha $