Exams Joint Entrance Examination Mathematics Hyperbola Question 32 Question 32 of 106 Medium NUMERICAL
Let H 1 : x 2 a 2 − y 2 b 2 = 1 {H}_{1}:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1 H 1 : a 2 x 2 − b 2 y 2 = 1 and H 2 : − x 2 A 2 + y 2 B 2 = 1 {H}_{2}:-\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1 H 2 : − A 2 x 2 + B 2 y 2 = 1 be two hyperbolas having length of latus rectums 15 2 15\sqrt{2} 15 2 and 12 5 12\sqrt{5} 12 5 respectively. Let their ecentricities be e 1 = 5 2 {e}_{1}=\sqrt{\frac{5}{2}} e 1 = 2 5 and e 2 {e}_{2} e 2 respectively. If the product of the lengths of their transverse axes is 100 10 100\sqrt{10} 100 10 , then 25 e 2 2 25{e}_{2}^{2} 25 e 2 2 is equal to \_\_\_\_\_\_\_\_\_ .
Answer: 55
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2 b 2 a = 15 2 1 + b 2 a 2 = 5 2 a = 5 2 b = 5 3 2 A 2 B = 12 5 2 a ⋅ 2 B = 100 10 2.5 2 .2 B = 100 10 B = 5 5 A = 5 6 e 2 2 = 1 + A 2 B 2 = 1 + 150 125 e 2 2 = 1 + 30 25 25 e 2 2 = 55 \begin{gathered} \frac{2{b}^{2}}{a}=15\sqrt{2} \\ 1+\frac{{b}^{2}}{{a}^{2}}=\frac{5}{2} \\ a=5\sqrt{2} \\ b=5\sqrt{3} \\ \frac{2{A}^{2}}{B}=12\sqrt{5} \\ 2a\cdot 2B=100\sqrt{10} \\ 2.5\sqrt{2}.2B=100\sqrt{10} \\ B=5\sqrt{5} \\ A=5\sqrt{6} \\ {e}_{2}^{2}=1+\frac{{A}^{2}}{{B}^{2}} \\ =1+\frac{150}{125} \\ {e}_{2}^{2}=1+\frac{30}{25} \\ 25{e}_{2}^{2}=55 \end{gathered} a 2 b 2 = 15 2 1 + a 2 b 2 = 2 5 a = 5 2 b = 5 3 B 2 A 2 = 12 5 2 a ⋅ 2 B = 100 10 2.5 2 .2 B = 100 10 B = 5 5 A = 5 6 e 2 2 = 1 + B 2 A 2 = 1 + 125 150 e 2 2 = 1 + 25 30 25 e 2 2 = 55 Answer the question to unlock