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Hyperbola question

2022 · 29 Jun · Shift 1 · Q40
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  5. /2022 · 29 Jun · Shift 1 · Q40

Hyperbola question

2022 · 29 Jun · Shift 1 · Q40

JEE MainMathematicsHyperbolaNumerical+4 / −1
Let H:x2a2−y2b2=1H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1H:a2x2​−b2y2​=1, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22+14)4(2\sqrt 2 + \sqrt {14} )4(22​+14​). If the eccentricity H is 112{{\sqrt {11} } \over 2}211​​, then the value of a2 + b2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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