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Let x^2 a^2+ y^2 b^2=1,a>b be an ellipse, whose eccentricity is 1 √2 and the length of the latus rectum is √14. Then the square of the eccentricity…

JEE Main 2024 — Mathematics Coordinate Geometry

2024mcqeasy

Let x2a2+y2b2=1,a>b\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,a>ba2x2​+b2y2​=1,a>b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}}2​1​ and the length of the latus rectum is 14\sqrt{14}14​. Then the square of the eccentricity of x2a2−y2b2=1\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1a2x2​−b2y2​=1 is:

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

Options

  1. A

    333

  2. B

    72\frac{7}{2}27​

  3. C

    32\frac{3}{2}23​

  4. D

    52\frac{5}{2}25​

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