Question

$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$.

Answer 1

$\begin{array}{c}\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2\to \left(i\right)\\ \Rightarrow \frac{x}{\cos \theta }=a,\frac{y}{\sin \theta }=b\\ \therefore x=a\cos \theta ,y=b\sin \theta \end{array}$

Put in equation (i), we get $\left(a^2-b^2\right)$.