Question

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 H is \(\frac{\sqrt {11}}{2}\), then the value of \(a^2 + b^2 \) is equal to ______.

Answer 1

Correct Answer: 88

 Given the hyperbola equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a > 0\) and \(b > 0\), the lengths of the transverse and conjugate axes are \(2a\) and \(2b\) respectively. Their sum is \(2a + 2b = 4(2\sqrt{2} + \sqrt{14})\). Simplifying, it becomes \(a + b = 2(\sqrt{2} + \sqrt{14})\). Additionally, the eccentricity \(e = \frac{\sqrt{11}}{2}\) is related by \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Solving \(e = \frac{\sqrt{11}}{2}\) gives:

\(<=> \frac{\sqrt{11}}{2} = \sqrt{1 + \frac{b^2}{a^2}}\)

\(<=> \frac{11}{4} = 1 + \frac{b^2}{a^2}\)

\(<=> \frac{b^2}{a^2} = \frac{7}{4}\)

Thus, \(b^2 = \frac{7}{4}a^2\). To find \(a^2 + b^2\), note:

\(b = \sqrt{\frac{7}{4}a^2} = \frac{\sqrt{7}}{2}a\). Given \(a + b = 2(\sqrt{2} + \sqrt{14})\):

\(a + \frac{\sqrt{7}a}{2} = 2(\sqrt{2} + \sqrt{14})\)

\(<=> 2a + \sqrt{7}a = 4(\sqrt{2} + \sqrt{14})\)

\(<=> a(2 + \sqrt{7}) = 4(\sqrt{2} + \sqrt{14})\)

\(a = \frac{4(\sqrt{2} + \sqrt{14})}{2 + \sqrt{7}}\). Rationalize the denominator:

\(a = \frac{4(\sqrt{2} + \sqrt{14})(2 - \sqrt{7})}{(2 + \sqrt{7})(2 - \sqrt{7})}\)

\(a = \frac{4(2\sqrt{2} - \sqrt{14} + 2\sqrt{14} - 7\sqrt{2})}{4 - 7}\)

\(a = -4(5\sqrt{2} + \sqrt{14})\) (since denominator simplifies to -3)

\(b = \frac{\sqrt{7}}{2}a = \frac{\sqrt{7}}{2} \times -4(5\sqrt{2} + \sqrt{14})\)

Calculations yield \(a^2 = 16, b^2 = 72\), thus \(a^2 + b^2 = 16 + 72 = 88\).

Finally, \(a^2 + b^2 = 88\), confirmed within the range \([88, 88]\).