Question

Locus of the point which divides double ordinates of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$ 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. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=a^2-b^2$

Answer 1

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

Given that : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let end points of double ordinates be $P(\alpha ,\beta )$ and $Q(\alpha ,-\beta )$
Let the point which divides $PQ$ in $1:2$ is $R(h,k)$
Then, by internal section formula
$h=\alpha$
And
$k=\frac{1\times \beta -2\times \beta }{3}\text{or}k=\frac{2\times \beta -1\times \beta }{3}\Rightarrow \beta =\pm 3k$

$\because (\alpha ,\beta )$ lies on ellipse
$\therefore \frac{h^2}{a^2}+\frac{(\pm 3k{)}^2}{b^2}=1$

Hence, the locus of $R$ is
$\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1$