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WBJEE · Maths · Ellipse

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

  1. A 9
  2. B 7
  3. C 11
  4. D 5
Verified Solution

Answer & Solution

Correct Answer

(C) 11

Step-by-step Solution

Detailed explanation

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…