JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{b^2}=1\) subtend an angle of \(\frac{\pi}{3}\) at the centre of the hyperbola. If \(\mathrm{b}^2\) is equal to \(\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})\), where \(l\) and \(\mathrm{m}\) are co-prime numbers, then \(l^2+\mathrm{m}^2+\mathrm{n}^2\) is equal to ...........
- A \(177\)
- B \(56\)
- C \(182\)
- D \(728\)
Answer & Solution
Correct Answer
(C) \(182\)
Step-by-step Solution
Detailed explanation
\( LR\) subtends \(60^{\circ}\) at centre \( \Rightarrow \tan 30^{\circ}=\frac{b^2 / a}{a e}=\frac{b^2}{a^2 e}=\frac{1}{\sqrt{3}} \) \( \Rightarrow \mathrm{e}=\frac{\sqrt{3}^2}{9}\) Also, \(e^2=1+\frac{b^2}{9} \Rightarrow 1+\frac{b^2}{9}=\frac{3 b^4}{81}\)…
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