Question

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

• A. 1
• B. 9
• C. 5
• D. 7

Answer 1

The correct option is D 7

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

$\Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$

$e_H^2=1+\frac{\frac{81}{25}}{\frac{144}{25}}=1+\frac{81}{144}$

$e_H^2=\frac{225}{149}$

$\Rightarrow e_H=\frac{15}{12}=\frac{5}{4}$

Foci of Hyperbola$(\pm \frac{12}{5}\cdot \frac{5}{4},0)\equiv (\pm 3,0)$

$\frac{x^2}{16}+\frac{y^2}{b^2}=1$ also have same foci

$\Rightarrow (\pm ae,0)=(\pm 3,0)$

$4e=3$

$\Rightarrow e=\frac{3}{4}$

$e^2=1-\frac{b^2}{16}=\frac{9}{16}$

$\frac{b^2}{16}=1-\frac{9}{16}=\frac{7}{16}$

$b^2=7$.