JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 13)
Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $${e^2} = {{11} \over {14}}l$$ and $${\left( {e'} \right)^2} = {{11} \over 8}l'$$, then the value of $$77a + 44b$$ is equal to :
100
110
120
130
Explanation
$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, then
$${e^2} = {{11} \over {14}}l$$ (l be the length of LR)
$$ \Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a$$ ..... (i)
and $$e{'^2} = {{11} \over 8}l'$$ (l' be the length of LR of conjugate hyperbola)
$$ \Rightarrow {a^2} + {b^2} = {{11} \over 4}{a^2}b$$ ....... (ii)
By (i) and (ii)
$$7a = 4b$$
then by (i)
$${{16} \over {49}}{b^2} + {b^2} = {{11} \over 7}{b^2}\,.\,{{4b} \over 7}$$
$$ \Rightarrow 44b = 65$$ and $$77a = 65$$
$$\therefore$$ $$77a + 44b = 130$$
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