JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 23)
Let the hyperbola $$H:{{{x^2}} \over {{a^2}}} - {y^2} = 1$$ and the ellipse $$E:3{x^2} + 4{y^2} = 12$$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If $${e_H}$$ and $${e_E}$$ are the eccentricities of H and E respectively, then the value of $$12\left( {e_H^2 + e_E^2} \right)$$ is equal to ___________.
Answer
42
Explanation
$$\because$$ $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 1} = 1$$
$$\therefore$$ Length of latus rectum $$ = {2 \over a}$$
$$E:{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$
Length of latus rectum $$ = {6 \over 2} = 3$$
$$\because$$ $${2 \over a} = 3 \Rightarrow a = {2 \over 3}$$
$$\therefore$$ $$12\left( {e_H^2 + e_E^2} \right) = 12\left( {1 + {9 \over 4}} \right) + \left( {1 - {3 \over 4}} \right) = 42$$
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