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

Let \(H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1\) be the hyperbola, whose eccentricity is \(\sqrt{3}\) and the length of the latus rectum is \(4 \sqrt{3}\). Suppose the point \((\alpha, 6), \alpha>0\) lies on \(H\). If \(\beta\) is the product of the focal distances of the point \((\alpha, 6)\), then \(\alpha^2+\beta\) is equal to :

  1. A \(170\)
  2. B \(171\)
  3. C \(169\)
  4. D \(172\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(171\)

Step-by-step Solution

Detailed explanation

\( \mathrm{H}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1, \mathrm{e}=\sqrt{3} \) \( \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{3} \quad \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 \) \( a^2=2 b^2 \)…
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