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The foci of the ellipse x^2 /16 +y^2/b^2 =1 and the hyperbola x^2/144-y^2 /81 =1/25 coincide , then the value of b^2 is

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Solution:

Foci of hyperbola $=(pm ae,0)=(pm 3,0)$

In Ellipse Foci $=(pm 4e,0)$ , Where $e=frac14sqrt16-b^2$

$herefore$ $sqrt16-b^2=3Rightarrow b^2=7$

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Deependra Verma

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