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
- A 9
- B 7
- C 11
- D 5
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…
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