JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(\mathrm{H}_1: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1\) and \(\mathrm{H}_2:-\frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1\) be two hyperbolas having length of latus rectums \(15 \sqrt{2}\) and \(12 \sqrt{5}\) respectively. Let their ecentricities be \(e_1=\sqrt{\frac{5}{2}}\) and \(e_2\) respectively. If the product of the lengths of their transverse axes is \(100 \sqrt{10}\), then \(25 \mathrm{e}_2^2\) is equal to ________.
- A 50
- B 51
- C 52
- D 55
Answer & Solution
Correct Answer
(D) 55
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) \(\frac{2 b^2}{a}=15 \sqrt{2}\) ...(i) \(\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{5}{2}}\)....(ii) From (i) and (ii) \(a=5 \sqrt{2} \text { and } b^2=75\) \(\frac{x^2}{A^2}-\frac{y^2}{B^2}=-1\) \(\frac{2 A^2}{B}=12 \sqrt{5}\) ....(iii) Since,…
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