JEE MAIN - Mathematics (2003 - No. 50)
The foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is :
$$9$$
$$1$$
$$5$$
$$7$$
Explanation
$${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$
$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$
$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$
$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$
$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola
$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$
$$\therefore$$ $$e = {3 \over 4}$$
Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$
$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$
$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$
$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$
$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$
$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola
$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$
$$\therefore$$ $$e = {3 \over 4}$$
Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$
$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$
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