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 is

• A. $ \sqrt{7}$
• B. $ \sqrt{6}$
• C. $ \sqrt{5}$
• D. $ \sqrt{8}$

Answer 1

Answer: (A)

$\frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1 $ and $\frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25} $ have same foci.
$ \therefore \:\: (ae, 0)$ of ellipse =$ (a' e', 0)$ of hyperbola
i.e., $ae = a'e'$
$4\sqrt{1-\frac{b^{2}}{16}} = \frac{12}{5} \sqrt{\frac{81}{25}\times \frac{25}{144}+1} = \frac{12}{5} \times \sqrt{\frac{25}{16}}$
$ \Rightarrow \sqrt{1- \frac{b^{2}}{16}} = \frac{3}{4} \Rightarrow b = \sqrt{7}$