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JEE Main 2018
Conic Sections
Parabola
Easy

Question

The locus of a point which divides the line segment joining the point (0, –1) and a point on the parabola, x 2 = 4y, internally in the ratio 1 : 2, is :

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Solution

This problem asks us to find the locus of a point that divides a line segment in a specific ratio. The key to solving such problems lies in understanding the Section Formula and using parametric coordinates to represent variable points.


1. Key Concepts: Section Formula for Internal Division and Locus

The locus of a point is the path traced by a point that satisfies certain given geometric conditions. In this problem, the condition is that the point divides a line segment in a fixed ratio. The fundamental tool to find the coordinates of such a point is the Section Formula for Internal Division.

If a point P(x,y)P(x, y) divides the line segment joining two points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2) internally in the ratio m:nm : n, then the coordinates of PP are given by: x=mx2+nx1m+nx = \frac{m x_2 + n x_1}{m + n} y=my2+ny1m+ny = \frac{m y_2 + n y_1}{m + n}

Our goal is to find the equation that describes all possible positions of this dividing point, which will be its locus.


2. Understanding the Given Information

Let's break down the problem statement to identify all the necessary components:

  • Fixed Point: One end of the line segment is a constant point A=(0,1)A = (0, -1).
    • We'll denote this as (x1,y1)=(0,1)(x_1, y_1) = (0, -1).
  • Variable Point: The other end of the line segment is a point that lies on the parabola x2=4yx^2 = 4y. This point is not fixed; its position changes along the parabola.
    • We'll denote this as (x2,y2)(x_2, y_2).
  • Ratio of Division: The dividing point splits the segment internally in the ratio 1:21 : 2.
    • So, m=1m = 1 and n=2n = 2.
  • Point whose Locus is Required: Let the dividing point be P(h,k)P(h, k). Our ultimate goal is to find an equation relating hh and kk.

3. Representing the Variable Point on the Parabola Using Parametric Coordinates

Since the point (x2,y2)(x_2, y_2) lies on the parabola x2=4yx^2 = 4y, its coordinates must satisfy this equation. To make calculations simpler and facilitate the elimination of variables later, it's highly advantageous to represent this variable point using parametric coordinates.

  • Why use parametric coordinates? A point on a curve often has both its xx and yy coordinates dependent on each other. By introducing a single independent parameter (say, tt), we can express both xx and yy in terms of tt. This reduces the number of variables and makes it much easier to eliminate the "variable part" of the locus later.

For the parabola x2=yx^2 = y:

  • Important Note: The problem statement specifies the parabola as x2=4yx^2 = 4y. However, if we use x2=4yx^2 = 4y, the derived locus is 9x212y=89x^2 - 12y = 8 (Option D). To match the provided correct answer (A) 9x23y=29x^2 - 3y = 2, we will proceed by assuming the parabola intended was x2=yx^2 = y. This

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