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 $|P M|:|P N|$ 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
$4 x \sec \theta-3 y \operatorname{cosec} \theta=7$
This meets the coordinate axes at
$M\left(\frac{7}{4} \cos \theta, 0\right), N\left(0,-\frac{7}{3} \sin \theta\right)$
$\therefore P M^{2}=\left(4-\frac{7}{4}\right)^{2} \cos ^{2} \theta+9 \sin ^{2} \theta$
$=\frac{9}{16}\left(9 \cos ^{2} \theta+16 \sin ^{2} \theta\right)$
$P N^{2}=16 \cos ^{2} \theta+\left(3+\frac{7}{3}\right)^{2} \sin \theta$
$=\frac{16}{9}\left(9 \cos ^{2} \theta+16 \sin ^{2} \theta\right)$
$\therefore P M^{2}: P N^{2}=9^{2}: 16^{2}$
$\Rightarrow |P M|:|P N|=9: 16$