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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

माना अतिपरवलय \(\frac{x^2}{9}-\frac{y^2}{b^2}=1\) का नाभिलम्ब, अतिपरवलय के केन्द्र प़र \(\frac{\pi}{3}\) का कोण बनाता है। यदि \(\mathrm{b}^2\) बराबर \(\frac{\ell}{\mathrm{m}}(1+\sqrt{\mathrm{n}})\) है, जहाँ \(\ell\) तथा \(\mathrm{m}\) असहभाज्य संख्याएँ हैं, तो \(\ell^2+m^2+n^2\) = ...........

  1. A \(177\)
  2. B \(56\)
  3. C \(182\)
  4. D \(728\)
Verified Solution

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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