Question

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

• A. $\frac{x^2}{a^2}-\frac{9y^2}{b^2}=\frac{1}{9}$
• B. $\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$
• C. $\frac{9y^2}{-a^2}+\frac{9y^2}{b^2}=1$
• D. None of these

Answer 1

Answer: (B)

Let $P(a\cos \theta ,b\sin \theta ),Q(a\cos \theta ,-b\sin \theta )$
Given, $PR:RQ=1:2$

Let a point $R(h,k)$ divides the line joining the points $P$ and $Q$ internally in the ratio $1:2$.
$\therefore h=a\cos \theta$
$\Rightarrow \cos \theta =\frac{h}{a}$ ... (i)
and $k=\frac{b}{3}\sin \theta$
$\Rightarrow \sin \theta =\frac{3k}{b}$ ... (ii)
On squaring and adding Eqs. (i) and (ii), we get
$\frac{h^2}{a^2}+\frac{9k^2}{b^2}=1$
Hence, locus of $R$ is
$\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$