Medium INTEGER +4 / -1 PYQ · JEE Mains 2022

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 (integer) 88

Solution

<p>$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$ ...... (1)</p> <p>$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$ ....... (2)</p> <p>$\Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$ ....... (3)</p> <p>and $a + b = 4\sqrt 2 + 2\sqrt {14}$ ...... (4)</p> <p>By (3) and (4)</p> <p>$\Rightarrow a + {{\sqrt 7 } \over 2}a = 4\sqrt 2 + 2\sqrt {14}$</p> <p>$$ \Rightarrow {{a\left( {2 + \sqrt 7 } \right)} \over 2} = 2\sqrt 2 \left( {2 + \sqrt 7 } \right)$$</p> <p>$\Rightarrow a = 4\sqrt 2 \Rightarrow {a^2} = 32$ and ${b^2} = 56$</p> <p>$\Rightarrow {a^2} + {b^2} = 32 + 56 = 88$</p>

About this question

Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola

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