Question

Let $A$ be a vertex of the ellipse $S \equiv \frac{x^{2}}{4}+\frac{y^{2}}{9}-1=0$ and $F$ be a focus of the ellipse $S^{\prime}=\frac{x^{2}}{9}+\frac{y^{2}}{4}-1=0$. Let $P$ be a point on the major axis of the ellipse $S^{\prime}=0$, which divides $\overline{O F}$ in the ratio $2: 1(O$ is the origin $) .$ If thc lcngth of the chord of the cllipsc $S=0$ through $A$ and $P$ is $\frac{3 \sqrt{101}}{k}$, then $k=$

• A. 5
• B. 4
• C. 7
• D. 8

Answer 1

Answer: (C)

We have, $S \equiv \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\therefore \, A(0, \pm 3)$ (vertex $=(0, \pm b)$
Again,
$ S' \equiv \frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$ \therefore \, F(\pm \sqrt{5}, 0)$ (Focus = $(\pm a e, 0)$
Again $P$ is a point on the major axis of the ellipse $S'=0$, which divides of in the ratio $2: 1 .$
$\therefore \,P-\left(\frac{2 x \pm \sqrt{5}+1 \times 0}{2+1}, \frac{2 \times 0+1 \times 0}{2+1}\right)-\left(\frac{\pm 2 \sqrt{5}}{3}, 0\right)$
Now, equation of line passing through $A$ and $P$ is given by
$y-3=\frac{0-3}{\frac{2 \sqrt{5}}{3}-0}(x-0)$
$ \Rightarrow \,y=-\frac{9}{2 \sqrt{5}} x+3\,\,\,\, \ldots (i)$
Now, intersection point of line in Eq. (i) and $S=0$ are $(0,3)$ and $\left(\frac{30}{7 \sqrt{5}}, \frac{-6}{7}\right)$.
$\therefore $ Length of chord
$=\sqrt{\left(\frac{30}{7 \sqrt{5}}-0\right)^{2}+\left(3+\frac{6}{7}\right)^{2}}=\sqrt{\frac{900}{247}+\frac{729}{49}}$
$=\sqrt{\frac{4545}{245}}=\sqrt{\frac{909}{49}}=\frac{3 \sqrt{101}}{7}$
$\therefore \,k=7$