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

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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},$$ \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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