$\begin{array}{1 1}(A)\;2 \\(B)\;3 \\(C)\;4 \\(D)\;1 \end{array}$

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Equation of normal of the ellipse

$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(a \cos \theta, b \sin \theta)$

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

Let the normal of the ellipse

$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ at $(a \cos \theta, b \sin \theta)$

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

Let normal is passing through the given point $(h,k)$ then

$\large\frac{ah}{\cos \theta}-\frac{bk}{\sin \theta}$$=a^2-b^2$

Let normal is passing through the given point $(h,k)$ then

$\large\frac{ah}{\cos \theta}-\frac{bk}{\sin \theta}$$=a^2-b^2$

=> $2ah(1+ \tan ^2 \large\frac{\theta}{2}) $$\tan \large\frac{\theta}{2}$$-bk (1- \tan ^4 \large\frac{\theta}{2})$

=> $2(a^2-b^2) \tan \large\frac{\theta}{2} $$(1- \tan ^2 \large\frac{\theta}{2})$

Hence it is a four degree polynomial in $\tan \large\frac{\theta}{2}$ therefore it may have four real roots.

Hence C is the correct answer.

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