Question

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

• A. $8$
• B. $10$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

Given equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$
eccentricity $=e=\sqrt{1-\frac{b^2}{16}}$
foci: $\pm 4\sqrt{1-\frac{b^2}{16}}$
Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$
eccentricity $=e=\sqrt{1+\frac{81}{25}\times \frac{25}{144}}=\sqrt{1+\frac{81}{25}}$
$=\sqrt{\frac{225}{144}}=\frac{15}{12}$
foci: $\pm \frac{12}{5}\times \frac{15}{12}=\pm 3$
Since, foci of ellipse and hyperbola coincide
$\therefore \pm 4\sqrt{1-\frac{b^2}{16}}=\pm 3\Rightarrow b^2=7$