Q76 / 200 · WBJEE 2014 · Mathematics

Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$ Then, the radius of the circle with centre $(0, \sqrt{2})$ and passing through the foci of the ellipse is

  • A
    9
  • B
    7
  • Correct answer
    C
    11
  • D
    5

Solution

Given equation of ellipse is Here, $$ \begin{array}{r} \frac{x^{2}}{144}+\frac{y^{2}}{25}=1 \\ a^{2}=144 \text { and } b^{2}=25 \end{array} $$ Now. $$ e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}} $$ $$ =\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12} $$ Foci of an ellipse $=(\pm a e, 0)$ $$ \begin{array}{l} =\left(\pm 12 \times \frac{\sqrt{119}}{12}, 0\right) \\ =(\pm \sqrt{119}, 0) \end{array} $$ Since, the circle with centre $(0, \sqrt{2})$ and passing through foci $(\pm \sqrt{119}, 0)$ of the ellipse. $\therefore$ Radius of circle $=\sqrt{(\sqrt{119}-0)^{2}+(0-\sqrt{2})^{2}}$ $$ =\sqrt{119+2}=\sqrt{121}=11 $$

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

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