Browse Questions

# The number of normal that can be drawn from a point to a given ellipse is

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

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.