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JEE Main 2024
Conic Sections
Ellipse
Hard

Question

Let the eccentricity of the ellipse x2+a2y2=25a2{x^2} + {a^2}{y^2} = 25{a^2} be b times the eccentricity of the hyperbola x2a2y2=5{x^2} - {a^2}{y^2} = 5, where a is the minimum distance between the curves y = e x and y = log e x. Then a2+1b2{a^2} + {1 \over {{b^2}}} is equal to :

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Solution

This problem is a comprehensive test of concepts from conic sections (ellipse and hyperbola) and differential calculus (finding the minimum distance between curves). We will systematically break down the problem into logical steps, explaining each part in detail.

1. Key Concepts and Formulas

Before diving into the solution, let's recall the essential definitions and formulas:

  • Standard Form of Ellipse: For an ellipse x2A2+y2B2=1\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1:
    • If A2>B2A^2 > B^2, the major axis is along the x-axis, and eccentricity e=1B2A2e = \sqrt{1 - \frac{B^2}{A^2}}.
    • If $

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