Question

Locus of the point which divides 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. None of the above

Answer 1

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

R divides the line PQ in the ratio $1:2$ then,

$h=\frac{1\times a\cos \theta +2\times a\cos \theta }{3}$

$\Rightarrow h=a\cos \theta$

$\Rightarrow \frac{h}{a}=\cos \theta$ ------(1)

And, $k=\frac{1\times (-b\sin \theta )+2\times b\sin \theta }{3}$

$k=\frac{b\sin \theta }{3}$

$\Rightarrow \frac{3k}{b}=\sin \theta$ -----(2)

Squaring and adding equations 1 and 2 we get:

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