If foci of ellipse frac{{{x^2}}}{{16}} + frac{{{y^2}}}{{{b^2}}} = 1 and hyperbola frac{{{x^2}}}{{144

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10-2. Parabola, Ellipse, Hyperbola

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If 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 the value of ${b^2}$ is

A

$1$

B

$5$

C

$7$

D

$9$

(AIEEE-2003)

Solution

(c) Hyperbola is $\frac{{{x^2}}}{{144}} – \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$

$a = \sqrt {\frac{{144}}{{25}}} ,\,\,b = \sqrt {\frac{{81}}{{25}}} ,\,\,{e_1} = \sqrt {1 + \frac{{81}}{{144}}}$

$= \sqrt {\frac{{225}}{{144}}} = \frac{{15}}{{12}} = \frac{5}{4}$

Therefore, foci $ = (a{e_1},0) = \left( {\frac{{12}}{5}.\frac{5}{4},0} \right) = (3,\,0)$

Therefore, focus of ellipse $ = (4e,0)$ $i.e.$ $(3,\,0)$

==> $e = $$3\over4$

Hence ${b^2} = 16\left( {1 – \frac{9}{{16}}} \right) = 7$.

Standard 11

Mathematics

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