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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
$8$
$10$
$7$
$9$
Solution
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 ae = \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{{144}}{{25}}} = \sqrt {1 + \frac{{81}}{{144}}} $
$ = \sqrt {\frac{{225}}{{144}}} = \frac{{15}}{{12}}$
foci $: \pm ae = \pm \frac{{12}}{5} \times \frac{{15}}{{12}} = \pm 3$
Since, foci of ellipse and hyperbola coincide
$\therefore \pm 4\sqrt {1 – \frac{{{b^2}}}{{16}}} = \pm 3 \Rightarrow {b^2} = 7$
