If the foci of the ellipse frac{{{x^2}}} | vedclass

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Question

Given equation of ellipse is 

$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$

eccentricity $ = e = \sqrt {1 - \frac{{{b^2}}}{{16}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

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}} 	imes \frac{{144}}{{25}}}  = \sqrt {1 + \frac{{81}}{{144}}} $

                   $ = \sqrt {\frac{{225}}{{144}}}  = \frac{{15}}{{12}}$

foci $: \pm ae =  \pm \frac{{12}}{5} 	imes \frac{{15}}{{12}} =  \pm 3$

Since, foci of ellipse and hyperbola coincide

$	herefore  \pm 4\sqrt {1 - \frac{{{b^2}}}{{16}}}  =  \pm 3 \Rightarrow {b^2} = 7$

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

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