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: $\pm \frac{12}{5}\times\frac{15}{12}=\pm3$
Since, foci of ellipse and hyperbola coincide
$\therefore \pm4 \sqrt{1-\frac{b^{2}}{16}}=\pm3 \Rightarrow b^{2}=7$