Question

If the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}}$ = 1 and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$ coincide, then the value of $b^{2}$ is

• A. 1
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

7

Answer 2

For hyperbola, $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25}$

$A = \sqrt{\frac{144}{25}},B = \sqrt{\frac{81}{25}},e_{1} = \sqrt{1 + \frac{B^{2}}{A^{2}}} = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{5}{4}$Therefore foci = $( \pm ae_{1},0) = \left( \pm \frac{12}{5}.\frac{5}{4},0 \right) = ( \pm 3,0)$. Therefore foci of ellipse i.e., $( \pm 4e,0) = ( \pm 3,0)$ (For ellipse $a = 4$)

⇒ $e = \frac{3}{4},$ Hence $P(x_{1},y_{1})$