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 ______.
Correct Answer: 88
Solution and Explanation
\(2a+2b=4(2\sqrt2+\sqrt14)\ \ ...(1)\)
\(1+\frac{b^2}{a^2}=\frac{11}{4}\ \ ....(2)\)
\(⇒\frac{b^2}{a^2}=\frac{7}{4}\ \ ...(3)\)
\(a+b=4\sqrt2+2\sqrt\ \ ....(4)\)
By \((3)\) and \((4)\)
\(a+\frac {\sqrt 7}{2}a = 4\sqrt 2+2\sqrt {14}\)
\(\frac {a(a+\sqrt 7)}{2}= 2\sqrt 2(2+\sqrt 7)\)
\(a=4\sqrt2\)
\(⇒\)\(a^2 = 32\) and \(b^2 = 56\)
\(⇒ a^2 + b^2 = 32 + 56\)
\(⇒ a^2 + b^2= 88\)
So, the answer is \(88\).
Top Questions on Hyperbola
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- If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
