Question

If the normal at any point $P$ of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ meets the coordinate axes at $M$ and $N$ respectively, then $|PM|:|PN|$ equals

• A. $4:3$
• B. $16:9$
• C. $9:16$
• D. $3:4$

Answer 1

Answer: (C)

The equation of the normal at $P(\theta )$ on the ellipse is
$4x\sec \theta -3y\mathrm{cosec}\theta =7$
This meets the coordinate axes at
$M\left(\frac{7}{4}\cos \theta ,0\right),N\left(0,-\frac{7}{3}\sin \theta \right)$
$\therefore P{M}^2={\left(4-\frac{7}{4}\right)}^2{\cos }^2\theta +9{\sin }^2\theta$
$=\frac{9}{16}\left(9{\cos }^2\theta +16{\sin }^2\theta \right)$
$P{N}^2=16{\cos }^2\theta +{\left(3+\frac{7}{3}\right)}^2\sin \theta$
$=\frac{16}{9}\left(9{\cos }^2\theta +16{\sin }^2\theta \right)$
$\therefore P{M}^2:P{N}^2=9^2:16^2$
$\Rightarrow |PM|:|PN|=9:16$