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