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JEE Main 2024
Conic Sections
Parabola
Medium

Question

Let P be a variable point on the parabola y=4x2+1y = 4{x^2} + 1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is :

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Solution

This problem requires us to find the locus of a midpoint, which involves a systematic application of coordinate geometry concepts: parameterization of points, finding the foot of a perpendicular, using the midpoint formula, and finally eliminating the parameter.

1. Key Concepts and Formulas

  • Parametric Representation: A point on a curve can often be represented using a single parameter (e.g., tt). For a parabola y=ax2+cy = ax^2 + c, a point P can be (t,at2+c)(t, at^2+c).
  • Foot of the Perpendicular: The foot of the perpendicular from a point P(x1,y1)P(x_1, y_1) to a line ax+by+c=0ax + by + c = 0 is a point $Q(x_2, y_

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