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← JEE Main 2022 · 29 Jun · Shift 1
MathematicsInteger+4 / -1Question 28 of 90

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 28

Let H:x2a2−y2b2=1H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1, a>0a>0, b>0b>0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(22+14)4\left(2\sqrt2+\sqrt{14}\right). If the eccentricity H is 112\dfrac{\sqrt{11}}2, then the value of a2+b2a^2+b^2 is equal to ______.

Answer and worked solution

Official answer: 88

Shape. e2=114e^2=\frac{11}4 gives b=72ab=\frac{\sqrt7}2a. Then a(1+72)=2(22+14)=22(2+7)a\left(1+\frac{\sqrt7}2\right)=2\left(2\sqrt2+\sqrt{14}\right)=2\sqrt2(2+\sqrt7), so a=42a=4\sqrt2, b=214b=2\sqrt{14}.

a2+b2=32+56=88a^2+b^2=32+56=88.

Answer: 88

Topics:Hyperbola