Q96 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

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 $S^{\prime} \equiv \frac{x^2}{9}+\frac{y^2}{4}-1=0$. 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 the length of the chord of the ellipse $S=0$ through $A$ and $P$ is $\frac{3 \sqrt{101}}{k}$, then $k=$

  • A
    5
  • B
    4
  • Correct answer
    C
    7
  • D
    8

Solution

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

Hence, the correct answer is option: 'C'.

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