JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 17)
A hyperbola passes through the foci of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :
$${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$
$${{{x^2}} \over 9} - {{{y^2}} \over 16} = 1$$
$${{{x^2}} \over 9} - {{{y^2}} \over 25} = 1$$
x2 $$-$$ y2 = 9
Explanation
$${e_1} = \sqrt {1 - {{16} \over {25}}} = {3 \over 5}$$ foci ($$ \pm $$ae, 0)
Foci = ($$ \pm $$3, 0)
Let equation of hyperbola be $${{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1$$
Passes through ($$ \pm $$3, 0)
A2 = 9, A = 3, $${e_2} = {5 \over 3}$$
$${e_2}^2 = 1 + {{{B^2}} \over {{A^2}}}$$
$${{25} \over 9} = 1 + {{{B^2}} \over 9} \Rightarrow {B^2} = 16$$
Equation of the hyperbola
$${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$
Foci = ($$ \pm $$3, 0)
Let equation of hyperbola be $${{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1$$
Passes through ($$ \pm $$3, 0)
A2 = 9, A = 3, $${e_2} = {5 \over 3}$$
$${e_2}^2 = 1 + {{{B^2}} \over {{A^2}}}$$
$${{25} \over 9} = 1 + {{{B^2}} \over 9} \Rightarrow {B^2} = 16$$
Equation of the hyperbola
$${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$
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