QJEEM22Ma7773: JEE Main 2022 Mathematics - Class 11 JEEM22Ma7773 Solution: Let , a > 0, b > 0, be a hyperbola such that the sum of lengths …

$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$ ...... (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 ${b^2} = 56$

$\Rightarrow {a^2} + {b^2} = 32 + 56 = 88$