Question

The normal at $P\left(\theta \right)$ and D$(\theta +\frac{\pi }{2})$ meet the major axis of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at Q and R. Then $P{Q}^2+D{R}^2=$

• A. $b^2(1-e^2)(2-e{)}^2$
• B. $a^2(1-e^2)(2-e{)}^2$
• C. $a^2(1+e^2)(2+e^2)$
• D. $b^2(1+e^2)(2+e^2)$

Answer 1

The correct option is B $a^2(1-e^2)(2-e{)}^2$

Equation of the normal at $P\left(\theta \right)$ is $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2{e}^2$
This normal meets the major axis at Q = $(a{e}^2\cos \theta ,0)$
Equation of the normal at $D(\frac{\pi }{2}+\theta )$ ie $D(-a\sin \theta ,b\cos \theta )$ is $\frac{-ax}{\sin \theta }-\frac{by}{\cos \theta }=a^2{e}^2$
This normal meets the major axis at R $=(-a{e}^2\sin \theta ,0)$
$\therefore P{Q}^2+D{R}^2=\left[\right(a{e}^2\cos \theta -a\cos \theta {)}^2+b^2{\sin }^2\theta ]+\left[\right(a{e}^2\sin \theta -a\sin \theta {)}^2+b^2{\cos }^2\theta ]$
$=a^2(1-e^2)(2-e^2)$ [on simplification]