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 =$(a^{\prime }{e}^{\prime },0)$ of hyperbola
i.e., $ae=a^{\prime }{e}^{\prime }$
$4\sqrt{1-\frac{b^2}{16}}=\frac{12}{5}\sqrt{\frac{81}{25}\times \frac{25}{144}+1}=\frac{12}{5}\times \sqrt{\frac{25}{16}}$
$\Rightarrow \sqrt{1-\frac{b^2}{16}}=\frac{3}{4}\Rightarrow b=\sqrt{7}$