JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let an ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\), \(a < b\), pass through the point \((4, 3)\) and have eccentricity \(\dfrac{\sqrt{5}}{3}\). Then the length of its latus rectum is :
- A \(\dfrac{4\sqrt{5}}{3}\)
- B \(2\sqrt{5}\)
- C \(\dfrac{7\sqrt{5}}{3}\)
- D \(\dfrac{8\sqrt{5}}{3}\)
Answer & Solution
Correct Answer
(D) \(\dfrac{8\sqrt{5}}{3}\)
Step-by-step Solution
Detailed explanation
Given the equation of the ellipse is \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) with \(a < b\). The eccentricity is given by \(e = \dfrac{\sqrt{5}}{3}\). Since \(a < b\), the relation between \(a\), \(b\), and \(e\) is \(a^2 = b^2(1 - e^2)\). Substituting the value of \(e\):…
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