JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the eccentricity \(e\) of a hyperbola satisfy the equation \(6e^2 - 11e + 3 = 0\). If the foci of the hyperbola are \((3, 5)\) and \((3, -4)\), then the length of its latus rectum is :
- A \(\dfrac{11}{3}\)
- B \(\dfrac{17}{3}\)
- C \(\dfrac{15}{2}\)
- D \(\dfrac{17}{2}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{15}{2}\)
Step-by-step Solution
Detailed explanation
Given the equation for eccentricity: \(6e^2 - 11e + 3 = 0\) \(6e^2 - 9e - 2e + 3 = 0\) \((3e - 1)(2e - 3) = 0\) \(e = \dfrac{1}{3}\) or \(e = \dfrac{3}{2}\) For a hyperbola, \(e > 1\), so \(e = \dfrac{3}{2}\). The distance between the foci \((3, 5)\) and \((3, -4)\) is \(2ae\).…
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