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Q)

If the foci of the ellipse $\large\frac{x^2}{16}+\frac{y^6}{b^2}$$=1$ and the hyperbola. $\large\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is

$\begin{array}{1 1}(A)\;1 \\(B)\;5 \\(C)\;7 \\(D)\;9 \end{array}$

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A)
For hyperbola
$e^2=1+\large\frac{b^2}{a^2}$
$\quad=1+\large\frac{81}{144}$
$\quad=\large\frac{225}{144}$
$\therefore e=\large\frac{15}{12}=\frac{5}{4}$
i.e $e > 1$
Also $a^2=\large\frac{144}{25}$
Hence the foci are $(\pm ae,o)$
i.e $ \bigg(\pm \large\frac{12}{5},\frac{5}{4} ,0 \bigg)$$= (\pm 3,0)$
Now, the foci coincide therefore for ellipse,
$ae= 3 \; or \;a^2e^2=9$
(or) $a^2 \bigg(1-\large\frac{b^2}{a^2} \bigg) $$=9$
$a^2-b^2=9$
or $16-9=b^2$
So $b^2=7$
Hence C is the correct answer.
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