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

Answer 1

Answer: (B)

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

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 $k=\frac{b}{3} \sin \theta$
$\Rightarrow \sin \theta=\frac{3 k}{b}$ ... (ii)
On squaring and adding Eqs. (i) and (ii), we get
$\frac{h^{2}}{a^{2}}+\frac{9 k^{2}}{b^{2}}=1$
Hence, locus of $R$ is
$\frac{x^{2}}{a^{2}}+\frac{9 y^{2}}{b^{2}}=1$