Question

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)

Given equation of ellipse is
$\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$
Here, $ a^{2}=16 $
$\Rightarrow a=4$
$\therefore e=\sqrt{1-\frac{b^{2}}{16}}=\frac{\sqrt{16-b^{2}}}{4}$
$\therefore $ Foci of ellipse are $(\pm a e, 0) i e,\left(\pm \sqrt{16-b^{2}}, 0\right)$
Also, given equation of hyperbola is
$ \frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$
Here, $ a^{2} =\left(\frac{12}{5}\right)^{2}, b^{2}=\left(\frac{9}{5}\right)^{2}$
$\therefore e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{5}{4}$
$\therefore $ Foci of the hyperbola are $(\pm a e, 0) i e,(\pm 3,0)$
According to the given condition, foci of ellipse = foci of hyperbola
$\therefore \sqrt{16-b^{2}}=3$
$ \Rightarrow b^{2}=7$