Exams Joint Entrance Examination Mathematics Hyperbola Question 53 Question 53 of 106 Medium NUMERICAL
Let H : x 2 a 2 − y 2 b 2 = 1 H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 H : a 2 x 2 − b 2 y 2 = 1 , a > 0 , b > 0 a>0,b>0 a > 0 , b > 0 , be a hyperbola such that the sum of the lengths of the transverse and conjugate axes is 4 ( 2 2 + 14 ) 4(2\sqrt{2}+\sqrt{14}) 4 ( 2 2 + 14 ) . If the eccentricity of H H H is 11 2 \dfrac{\sqrt{11}}2 2 11 , then the value of a 2 + b 2 a^2+b^2 a 2 + b 2 is equal to:
Answer: 88
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2 a + 2 b = 4 ( 2 2 + 14 ) ⇒ a + b = 4 2 + 2 14 2a+2b=4(2\sqrt{2}+\sqrt{14})\Rightarrow a+b=4\sqrt{2}+2\sqrt{14} 2 a + 2 b = 4 ( 2 2 + 14 ) ⇒ a + b = 4 2 + 2 14 .
e 2 = 1 + b 2 a 2 = 11 4 ⇒ b 2 a 2 = 7 4 ⇒ b = 7 2 a e^2=1+\dfrac{b^2}{a^2}=\dfrac{11}4\Rightarrow\dfrac{b^2}{a^2}=\dfrac74\Rightarrow b=\dfrac{\sqrt{7}}2a e 2 = 1 + a 2 b 2 = 4 11 ⇒ a 2 b 2 = 4 7 ⇒ b = 2 7 a .
\[a+\frac{\sqrt{7}}2a=4\sqrt{2}+2\sqrt{14}\ \Longrightarrow\ \frac{a(2+\sqrt{7})}2=2\sqrt{2}(2+\sqrt{7})\ \Longrightarrow\ a=4\sqrt{2}.\]
\[a^2=32,\qquad b^2=\frac74\cdot32=56,\qquad a^2+b^2=88.\] Answer the question to unlock