JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 21)
Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $$4(2\sqrt 2 + \sqrt {14} )$$. If the eccentricity H is $${{\sqrt {11} } \over 2}$$, then the value of a2 + b2 is equal to __________.
Answer
88
Explanation
$$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$$ ...... (1)
$$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$$ ....... (2)
$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$$ ....... (3)
and $$a + b = 4\sqrt 2 + 2\sqrt {14} $$ ...... (4)
By (3) and (4)
$$ \Rightarrow a + {{\sqrt 7 } \over 2}a = 4\sqrt 2 + 2\sqrt {14} $$
$$ \Rightarrow {{a\left( {2 + \sqrt 7 } \right)} \over 2} = 2\sqrt 2 \left( {2 + \sqrt 7 } \right)$$
$$ \Rightarrow a = 4\sqrt 2 \Rightarrow {a^2} = 32$$ and $${b^2} = 56$$
$$ \Rightarrow {a^2} + {b^2} = 32 + 56 = 88$$
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