JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
For some \(\theta\in(0,\frac{\pi}{2}),\) let the eccentricity and the length of the latus rectum of the hyperbola \(x^{2}-y^{2}sec^{2}\theta=8\) be \(e_{1}\) and \(l_{1}\), respectively, and let the eccentricity and the length of the latus rectum of the ellipse \(x^{2}sec^{2}\theta+y^{2}=6\) be \(e_{2}\) and \(l_{2},\) respectively. If \(e_{1}^{2}=e_{2}^{2}(sec^{2}\theta+1)\), then \((\frac{l_{1}l_{2}}{e_{1}e_{2}})tan^{2}\theta\) is equal to ___ .
- A 4
- B 6
- C 8
- D 10
Answer & Solution
Correct Answer
(C) 8
Step-by-step Solution
Detailed explanation
\(\frac{x^2}{8}-\frac{y^2}{8 \cos ^2 \theta}=1, e_1=\sqrt{1+\frac{8 \cos ^2 \theta}{8}}\) \(\ell_1=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \cdot\left(8 \cos ^2 \theta\right)}{2 \sqrt{2}}\)…
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