Question

If the line $x \cos \alpha+y \sin \alpha=p$ be normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then

(A) $p^2\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right)=a^2-b^2$

(B) $p^2\left(a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha\right)=\left(a^2-b^2\right)^2$

(C) $p^2\left(a^2 \sec ^2 \alpha+b^2 \operatorname{cosec}^2 \alpha\right)=a^2-b^2$

(D) $p^2\left(a^2 \sec ^2 \alpha+b^2 \operatorname{cosec}^2 \alpha\right)=\left(a^2-b^2\right)^2$

Answer 1

At point $(a \cos \theta, b \sin \theta)$, equation of normal is $a x \sec \theta-b y \operatorname{cosec} \theta=a^2-b^2$

If this is to be same as the line $x \cos \alpha+y \sin \alpha=p$, then

$\frac{a \sec \theta}{\cos \alpha}=\frac{-b \operatorname{cosec} \theta}{\sin \alpha}=\frac{a^2-b^2}{p}$

$\Rightarrow \quad \cos \theta=\frac{a p}{\cos \alpha\left(a^2-b^2\right)}$

$\sin \theta=\frac{-b p}{\sin \alpha\left(a^2-b^2\right)}$

$\therefore \quad \cos ^2 \theta+\sin ^2 \theta=\frac{a^2 p^2}{\cos ^2 \alpha\left(a^2-b^2\right)^2}+\frac{b^2 p^2}{\sin ^2 \alpha\left(a^2-b^2\right)^2}=1$

$\Rightarrow \quad p^2\left(a^2 \sec ^2 \alpha+b^2 \operatorname{cosec}^2 \alpha\right)=\left(a^2-b^2\right)^2$

So, The correct option will be (D).