If the polar of $y^2=4ax$ is always touching the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ then locus of the pole is :

Question

$\begin{array}{1 1}(A)\;4a^2x^2-b^2y^2=4a^4 \\(B)\;a^2x^2-b^2y^2=a^4 \\(C)\;4a^2x^2+b^2y^2=4a^4 \\(D)\;none\;of\;these \end{array}$