Question

If the normal at any point $P$ on the ellipse $\frac{x^2}{9}+\frac{y^2}{16}=1$ meets the axes at $G$ and $g,$ respectively, then the ratio of $PG:Pg$ is

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

Answer 1

The correct option is D $16:9$

Let $P(a\cos \theta ,b\sin \theta )$ be a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$
Then the equation of the normal at $P$ is $ax\sec \theta -by\text{cosec}\theta =a^2-b^2.$
It meets the axes at $G(\frac{a^2-b^2}{a}\cos \theta ,0)$ and $g(0,-\frac{a^2-b^2}{a}\sin \theta )$

$\therefore P{G}^2={(a\cos \theta -\frac{a^2-b^2}{a}\cos \theta )}^2+(b\sin \theta -0{)}^2$
$=\frac{b^2}{a^2}(b^2{\cos }^2\theta +a^2{\sin }^2\theta )$
$P{g}^2=\frac{a^2}{b^2}(b^2{\cos }^2\theta +a^2{\sin }^2\theta )$

$\Rightarrow P{G}^2:P{g}^2=b^4:a^4$
$\Rightarrow PG:Pg=b^2:a^2$
$\Rightarrow PG:Pg=16:9$