If $x \cos \alpha +y \sin \alpha =x \cos \beta +y \sin \beta =2a\text{, }$then $\cos \alpha \cos \beta$ is,

Solution:

We have,

$x\cos \alpha +y\sin \alpha =x\cos \beta +y\sin \beta =2a$

$\Rightarrow \alpha ,\beta$ are the roots of $x \cos \theta +y \sin \theta =2a\text{. }$

Now, 

$\begin{array}{l}x\cos \theta +y\sin \theta =2a\\ \Rightarrow (x\cos \theta -2a{)}^2=y^2{\sin }^2\theta \\ \Rightarrow x^2{\cos }^2\theta -4ax\cos \theta +4a^2=y^2\left(1-{\cos }^2\theta \right)\\ \Rightarrow \left(x^2+y^2\right){\cos }^2\theta -4ax\cos \theta +4a^2-y^2=0\end{array}$

Clearly, $\cos \alpha ,\cos \beta$ are the roots of this equation. 

$\therefore \cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$.