$\begin{array}{1 1}(A)\;a^2y^2+b^2x^2 =4x^2y^2 \\(B)\;a^2x^2+b^2y^2=4x^2y^2 \\(C)\;x^2+y^2=a^2 \\(D)\;x^2+y^2=b^2 \end{array}$

Any tangent to the ellipse is $\large\frac{x}{a}$$ \cos \theta+ \large\frac{y}{b}$$ \sin \theta=1$

It meets the axes at $A \bigg( \large\frac{a}{\cos \theta}, 0 \bigg)$$ B\bigg( 0, \large\frac{b}{\sin \theta} \bigg)$

If $(h,k)$ be the mid - point of AB then

$2h=\large\frac{a}{\cos \theta}$

$2k=\large\frac{b}{\sin \theta}$

$\therefore \cos \theta=\large\frac{a}{2h} \qquad $$\sin \theta =\large\frac{b}{2k}$

$\therefore \cos ^2 \theta +\sin ^2 \theta=1$

$\therefore \large\frac{a^2}{4h^2} +\frac{b^2}{4k^2}$$=1$

or $a^2y^2+b^2x^2=4x^2y^2$

Hence A is the correct answer.

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