Question

If the normal at any point P on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the co-ordinate axes in G and g respectively, then PG : Pg =

• A. $a:b$
• B. $a^2:b^2$
• C. $b^2:a^2$
• D. $b:a$

Answer 1

The correct option is C $b^2:a^2$

Let $P(acos\theta ,bsin\theta )$ 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 $axsec\theta -bycosec\theta =a^2-b^2$
It meets the co-ordinate axes at $G(\frac{a^2-b^2}{a}cos\theta ,0)$ and $g(0,-\frac{a^2-b^2}{b}sin\theta ).$
$\Rightarrow P{G}^2={(acos\theta -\frac{a^2-b^2}{a}cos\theta )}^2+b^{2}si{n}^2\theta$
$=\frac{b^2}{a^2}(b^{2}co{s}^2\theta +a^{2}si{n}^2\theta )$
and $P{g}^2=\frac{a^2}{b^2}(b^{2}co{s}^2\theta +a^{2}si{n}^2\theta ),\therefore PG:pg=b^2:a^2$.