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JEE Main 2023
Conic Sections
Parabola
Hard

Question

Let the tangent to the parabola y2=12x\mathrm{y}^{2}=12 \mathrm{x} at the point (3,α)(3, \alpha) be perpendicular to the line 2x+2y=32 x+2 y=3. Then the square of distance of the point (6,4)(6,-4) from the normal to the hyperbola α2x29y2=9α2\alpha^{2} x^{2}-9 y^{2}=9 \alpha^{2} at its point (α1,α+2)(\alpha-1, \alpha+2) is equal to _________.

Answer: 3

Solution

This problem is a comprehensive test of your understanding of conic sections, specifically parabolas and hyperbolas, along with fundamental concepts of coordinate geometry like tangents, normals, perpendicular lines, and distance formulas. We will break down the problem into logical steps, applying relevant formulas and concepts at each stage to determine the square of the required distance.


Problem Overview

Our goal is to find the square of the distance of a given point (6,4)(6,-4) from a specific normal to a hyperbola. To achieve this, we must first:

  1. Determine the value of α\alpha: This involves using the properties of the parabola y2=12xy^2=12x and its tangent at (3,α)(3, \alpha), along with the condition that this tangent is perpendicular to the line

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