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

Answer 1

The correct option is B

$\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$

Explanation for the correct option:
Given the equation of ellipse as

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Using he ratio we get

$PR:PQ=1:2$

Then the value of $h$ is obtains as

$$\begin{gathered} \Rightarrow h=a\cos \theta \\ \Rightarrow \frac{h}{a}=\cos \theta \end{gathered}$$

Similarly,

$$\begin{gathered} \Rightarrow k=\frac{2b\sin \theta +1(-b\sin \theta )}{3} \\ \Rightarrow k=\frac{b\sin \theta }{3} \\ \Rightarrow \sin \theta =\frac{3k}{b} \end{gathered}$$

Consider, $\frac{h}{a}=\cos \theta$ and squaring both sides

$\Rightarrow {\cos }^2\theta =\frac{h^2}{a^2}.....\left(i\right)$

Similarly, consider $\sin \theta =\frac{3k}{b}$and squaring both sides

$\Rightarrow {\sin }^2\theta =\frac{9k^2}{b^2}....\left(ii\right)$

Now add equations $\left(i\right)\&\left(ii\right)$

$$\begin{gathered} \Rightarrow \frac{h^2}{a^2}+\frac{9k^2}{b^2}={\cos }^2\theta +{\sin }^2\theta \\ \Rightarrow \frac{h^2}{a^2}+\frac{9k^2}{b^2}=1 \end{gathered}$$

Hence, option (B) is the correct answer.