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JEE Main 2024
Conic Sections
Hyperbola
Hard

Question

Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1,a>bbe(±5,0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0) and 50\sqrt{50}, respectively. Then, the square of the eccentricity of the hyperbola x2b2y2a2b2=1\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1 equals

Answer: 5

Solution

This problem requires us to first extract parameters from a given ellipse and then use these parameters to determine the eccentricity of a related hyperbola. We will use the standard definitions and formulas for ellipses and hyperbolas.

1. Key Concepts and Formulas

Before diving into the solution, let's recall the essential properties of ellipses and hyperbolas that will be used:

  • Ellipse (Major axis along x-axis): The standard equation is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, where a>ba>b.
    • Foci: (±ae,0)(\pm ae, 0), where ee is the eccentricity.
    • Length of the latus rectum: $\frac{

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