If  ${\cos }^{-1}\left(\frac{p}{a}\right)+{\cos }^{-1}\left(\frac{q}{b}\right)=\alpha$, then $\frac{p^2}{a^2}+k\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha$, where ‘k’ is equal to

${\cos }^{-1}\left[\frac{p}{a}\cdot \frac{q}{b}-\sqrt{(1-\frac{p^2}{a^2})}\sqrt{(1-\frac{q^2}{b^2})}\right]=\alpha$

$\Rightarrow \frac{pq}{ab}-\sqrt{(1-\frac{p^2}{a^2})}\sqrt{(1-\frac{q^2}{b^2})}=\cos \alpha$

$\Rightarrow {(\frac{pq}{ab}-\cos \alpha )}^2=1-\frac{p^2}{a^2}-\frac{q^2}{b^2}+\frac{p^2{q}^2}{a^2{b}^2}$

$\Rightarrow \frac{p^2{q}^2}{a^2{b}^2}+{\cos }^2\alpha -\frac{2pq}{ab}\cos \alpha =1-\frac{p^2}{a^2}-\frac{q^2}{b^2}+\frac{p^2{q}^2}{a^2{b}^2}$

$\Rightarrow \frac{p^2}{a^2}-\frac{2pq}{ab}\cos \alpha +\frac{q^2}{b^2}=1-{\cos }^2\alpha ={\sin }^2\alpha$

$\therefore k=\frac{-2pq}{ab}$