Question

Locus of the point which divided the double ordinates of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in the ratio $1:2$ internally is

• A. $\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$
• B. $\frac{x^2}{a^2}+\frac{9y^2}{b^2}=\frac{1}{9}$
• C. $\frac{9x^2}{a^2}+\frac{9y^2}{b^2}=1$
• D. $\frac{9x^2}{a^2}+\frac{9y^2}{b^2}=\frac{1}{9}$

Answer 1

The correct option is A $\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$


Let
$P\equiv (a\cos \theta ,b\sin \theta )$, $Q\equiv (a\cos \theta ,-b\sin \theta )$.
$PR:RQ=1:2$
$\therefore h=a\cos \theta$
$\Rightarrow \cos \theta =\frac{h}{a}\cdots \left(i\right)$
$k=\frac{b}{3}\sin \theta$
$\Rightarrow \sin \theta =\frac{3k}{b}\cdots \left(ii\right)$
On squaring and adding (i) and (ii), we get
$\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$