Question

Locus of the middle points of all chords of $\frac{x^2}{4}+\frac{y^2}{9}=1$, which are at a distance of $2$ units from the vertex of parabola $y^2=-8ax$ is

• A. ${\left(\frac{x^2}{4}+\frac{y^2}{9}\right)}^2=\frac{xy}{6}$
• B. ${\left(\frac{x^2}{4}+\frac{y^2}{9}\right)}^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)$
• C. ${\left(\frac{x^2}{4}+\frac{y^2}{9}\right)}^2=\frac{x^2}{9}+\frac{y^2}{4}$
• D. None of the above

Answer 1

Answer: (B)

${\left(\frac{x^2}{4}+\frac{y^2}{9}\right)}^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)$

Let (h, k) be the mid-point of a chord of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$. Then, its equation is
$\frac{hx}{4}+\frac{ky}{9}-1=\frac{h^2}{4}+\frac{k^2}{9}-1$ $[\because T=S_1]$
$\Rightarrow \frac{hx}{4}+\frac{ky}{9}=\frac{h^2}{4}+\frac{k^2}{9}$
Since, it is a distance of $2$ units from the vertex $(0,0)$ of the parabola $y^2=-8ax$.
Therefore, $\left|\frac{\frac{h^2}{4}+\frac{k^2}{9}}{\sqrt{\frac{h^2}{16}+\frac{k^2}{81}}}\right|=2$
$\Rightarrow {\left(\frac{h^2}{4}+\frac{k^2}{9}\right)}^2=4\left(\frac{h^2}{16}+\frac{k^2}{81}\right)$
Hence, the locus of $(h,k)$ is
${\left(\frac{x^2}{4}+\frac{y^2}{9}\right)}^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)$.