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JEE Main 2023
Conic Sections
Ellipse
Hard

Question

Let the tangent and normal at the point (33,1)(3 \sqrt{3}, 1) on the ellipse x236+y24=1\frac{x^{2}}{36}+\frac{y^{2}}{4}=1 meet the yy-axis at the points AA and BB respectively. Let the circle CC be drawn taking ABA B as a diameter and the line x=25x=2 \sqrt{5} intersect CC at the points PP and QQ. If the tangents at the points PP and QQ on the circle intersect at the point (α,β)(\alpha, \beta), then α2β2\alpha^{2}-\beta^{2} is equal to :

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Solution

This problem is a comprehensive test of concepts involving ellipses, tangents, normals, circles, and poles/polars. We will systematically break down the problem into several parts to ensure clarity and accuracy.

1. Finding the Equation of the Tangent to the Ellipse and its y-intercept A

  • Key Concept: The equation of the tangent to the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 at a point (x1,y1)(x_1, y_1) is given by the formula T=0T=0, which is xx1a2+yy1b2=1\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1. This formula is a fundamental result

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