Question

Let $L$ is distance between two parallel normals of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b,$ then maximum value of $L$ is:-

• A. $2a$
• B. $2b$
• C. $2(a-b)$
• D. $2(a+b)$

Answer 1

The correct option is C $2(a-b)$

Equation of the normal to the given ellipse at $(a\cos \theta ,b\sin \theta )$ is
$ax\sec \theta -by\text{cosec}\theta =a^2-b^2$
$\because$ lines are parallel
$\therefore m_1=m_2{m}_1=\tan \theta ,m_2=\tan (\theta +\pi )$
$\Rightarrow ax\sec \theta -by\text{cosec}\theta =-(a^2-b^2)$
$L=\frac{2(a^2-b^2)}{\sqrt{a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta }}$
$a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta \ge (a+b{)}^2$
$L\le 2(a-b)$