Question

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

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

Answer 1

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

Let $P\equiv (a\cos \theta ,b\sin \theta )$
Thus equation of normal to the given ellipse at 'P' is given by,
$ax\sec \theta -bycosec\theta =a^2-b^2$
$\therefore G\equiv \left(\right(a-\frac{b^2}{a})\cos \theta ,0),g\equiv (0,(b-\frac{a^2}{b}\left)\sin \theta \right)$
Thus $PG=\sqrt{\frac{b^4}{a^2}{\cos }^2\theta +b^2{\sin }^2\theta }=\frac{b}{a}\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
and $Pg=\sqrt{a^2{\cos }^2\theta +\frac{a^4}{b^2}{\sin }^2\theta }=\frac{a}{b}\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
$\therefore PG:Pg=\frac{b^2}{a^2}$