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 \underline\hspace2cm .

Hyperbola PYQ — JEE Main 2022 · 29 Jun · Shift 1 · Q40 with Solution

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Correct answer: 88

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ the lengths of the transverse and conjugate axes are: $2a \quad \text{and} \quad 2b.$
Their sum is given as $2a+2b=4(2\sqrt{2}+\sqrt{14}).$ Dividing by $2$ : $a+b=2(2\sqrt{2}+\sqrt{14})=4\sqrt{2}+2\sqrt{14}.$
For this hyperbola, eccentricity satisfies $e^2=1+\frac{b^2}{a^2}.$ Given $e=\frac{\sqrt{11}}{2},$ so $\frac{11}{4}=1+\frac{b^2}{a^2}.$ Hence, $\frac{b^2}{a^2}=\frac{11}{4}-1=\frac{7}{4}.$ Therefore, $\frac{b}{a}=\frac{\sqrt{7}}{2}$ (since $a,b>0$ ), so $b=\frac{\sqrt{7}}{2}a.$
Substitute into the sum relation: $a+b=a+\frac{\sqrt{7}}{2}a=a\left(1+\frac{\sqrt{7}}{2}\right)=4\sqrt{2}+2\sqrt{14}.$

Note that $4\sqrt{2}+2\sqrt{14}=2\sqrt{2}(2+\sqrt{7}).$ Also, $1+\frac{\sqrt{7}}{2}=\frac{2+\sqrt{7}}{2}.$ So $a\cdot \frac{2+\sqrt{7}}{2}=2\sqrt{2}(2+\sqrt{7}).$ Cancelling $(2+\sqrt{7})$ : $a=4\sqrt{2}.$
Then $b=\frac{\sqrt{7}}{2}\cdot 4\sqrt{2}=2\sqrt{14}.$
Now compute: $a^2+b^2=(4\sqrt{2})^2+(2\sqrt{14})^2=32+56=88.$

Therefore, the required value is $\boxed{88}.$