Question

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

• A. $4$
• B. $9$
• C. $7$
• D. none

Answer 1

Answer: (C)

$7$

For hyperbola

$e^2=1+\frac{b^2}{a^2}$

$=1+\frac{9^2}{{12}^2}$

$=1+\frac{81}{144}=\frac{225}{144}$

$\therefore e=\frac{15}{12}=\frac{5}{4}$

i.e $e>1$

Also,$a^2=\frac{144}{25}$

Hence the foci are $(\pm ae,0)$

$=(\pm \frac{12}{5}\times \frac{5}{4},0)$

$=(\pm 3,0)$

Now, the foci coincide therefore for ellipse,

$ae=3$ or $a^2{e}^2=9$

or $a^2(1-\frac{b^2}{a^2})=9$

$\Rightarrow a^2-b^2=9$

or $16-b^2=9$

So,$b^2=16-9=7$