Question

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide then $b^2=$ ?

• A. 5
• B. 7
• C. 9
• D. 1

Answer 1

The correct option is B 7

Given hyperbola is, $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{x^2}{144/25}-\frac{y^2}{81/25}=1$
$\Rightarrow a^2=\frac{144}{25},b^2=\frac{81}{25},e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}$
$\therefore$ foci of hyperbola are $(\pm ae,0)i.e(\pm 3,0)$
Now given ellipse is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$
$\Rightarrow a^2=16$
Assume eccentricity of this ellipse is $e^{\prime }$ then its foci are $(\pm a{e}^{\prime },0)i.e(\pm 4e^{\prime },0)$
Given focci of given hyperbola and ellipse coincide
$\Rightarrow 4e^{\prime }=3\Rightarrow e^{\prime }=\frac{3}{4}$
For ellipse, using eccentricity relationship, $e^{\prime 2}=1-\frac{b^2}{a^2}$
$\Rightarrow \frac{9}{16}=1-\frac{b^2}{16}$
$\therefore b^2=7$