Question

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

• A. 3
• B. 16
• C. 9
• D. 12

Answer 1

Answer: (B)

16

Answer 2

If eccentricities of ellipse and hyperbola are e & $e_{1}\therefore$foci $( \pm ae,0)$ & $\left( \pm a_{1}e_{1},0 \right)$

Here $ae = a_{1}e_{1}$

$a^{2}e^{2} = a_{1}^{2}e_{1}^{2}$

$a^{2}\left( 1 - \frac{b^{2}}{a^{2}} \right) = a_{1}^{2}\left( 1 - \frac{b_{1}^{2}}{a_{1}^{2}} \right)$

⇒ $a^{2} - b^{2} = a_{1}^{2} + b_{1}^{2}$

⇒ $25 - b^{2} = \frac{144}{25} + \frac{81}{25}$= 9

$\therefore$ $b^{2} = 16$