Let H : frac{x^{2}}{a^{2}}-frac{y^{2}}{b^{2}} | vedclass

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(D) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Given the eccentricity $e = \frac{\sqrt{11}}{2}$,we have $e^{2} = 1 +

Vedclass · Accepted Answer

$88$

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(D) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Given the eccentricity $e = \frac{\sqrt{11}}{2}$,we have $e^{2} = 1 + \frac{b^{2}}{a^{2}}$.Substituting the value of $e$,we get $\frac{11}{4} = 1 + \frac{b^{2}}{a^{2}}$,which implies $\frac{b^{2}}{a^{2}} = \frac{7}{4}$,so $b = \frac{\sqrt{7}}{2}a$.The sum of the lengths of the transverse axis $(2a)$ and the conjugate axis $(2b)$ is given as $2a + 2b = 4(2\sqrt{2} + \sqrt{14})$.Substituting $b = \frac{\sqrt{7}}{2}a$,we get $2a + 2(\frac{\sqrt{7}}{2}a) = 4(2\sqrt{2} + \sqrt{14})$.$2a + \sqrt{7}a = 4(2\sqrt{2} + \sqrt{2}\sqrt{7})$.$a(2 + \sqrt{7}) = 4\sqrt{2}(2 + \sqrt{7})$.Thus,$a = 4\sqrt{2}$,which means $a^{2} = 16 	imes 2 = 32$.Since $b^{2} = \frac{7}{4}a^{2}$,we have $b^{2} = \frac{7}{4} 	imes 32 = 7 	imes 8 = 56$.Therefore,$a^{2} + b^{2} = 32 + 56 = 88$.

Let $H : \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,$a > 0, b > 0$,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 of $H$ is $\frac{\sqrt{11}}{2}$,then the value of $a^{2}+b^{2}$ is equal to

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