Problem 140 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(P\) be the point \((1,0)\) and \(Q\) a point on the locus \(y^{2}=8 x .\) The locus of mid point of \(P Q\) is (a) \(y^{2}-4 x+2=0\) (b) \(y^{2}+4 x+2=0\) (c) \(x^{2}+4 y+2=0\) (d) \(x^{2}-4 y+2=0\)

Step-by-Step Solution

Answer

None of the given equations match the locus of the midpoint; it's \(y^2 = 4x - 1\).

Step 1: Identify the Equation of the Locus

The equation of the locus is given as \(y^2 = 8x\). This is a standard parabola opening to the right.

Step 2: Determine Coordinates of the Midpoint

If \(Q\) is a point on the parabola \(y^2 = 8x\), let \(Q = (x_1, y_1)\). Then, the coordinates of the midpoint \(M\) of segment \(PQ\) are:\[M = \left( \frac{1 + x_1}{2}, \frac{0 + y_1}{2} \right)\]which simplifies to \(M = \left( \frac{1 + x_1}{2}, \frac{y_1}{2} \right)\).

Step 3: Express Midpoint Coordinates in Terms of Locus Equation

Since \(Q\) is on \(y^2 = 8x\), \(y_1^2 = 8x_1\). Substitute \(x_1 = 2x - \frac{1}{2}\) and \(y_1 = 2y\) into the equation:\[(2y)^2 = 8(2x - \frac{1}{2})\]

Step 4: Simplify the Equation

Simplify the equation from the previous step:\[4y^2 = 16x - 4\]Dividing throughout by 4, we get:\[y^2 = 4x - 1\]So, the equation of the locus of the midpoint is \(y^2 = 4x - 1\).

Step 5: Relate to Given Options

None of the given options are written as \(y^2 = 4x - 1\) directly. However, arranging \(y^2 - 4x + 2 = 0\) results in:\[y^2 = 4x - 2\]This shows that none of the listed options represent the correct equation we found for the locus of the midpoint.

Key Concepts

Locus

In mathematics, the term *locus* refers to a set of points that satisfy a particular condition or equation. When discussing parabolas or any curve, the locus is the collection of all points

• that satisfy the shape's equation.

For example, in the exercise we start with the equation of a parabola: \(y^2 = 8x\). This equation describes a locus that forms a parabolic curve opening to the right.
The points on this parabola are all coordinates \((x, y)\) that satisfy this equation. It's important to note that when a point moves along the locus, it adheres to the rule set by the equation.
Understanding the locus helps us predict how the parabola will behave, and which coordinates will form part of the curve's path.

Midpoint

The midpoint of a line segment is the point that divides the segment into two equal parts. To find the midpoint between two points in coordinate geometry,

• you average the x-coordinates and the y-coordinates of the two points.

This concept is pivotal in many geometric problems as it helps in figuring out the center or balance point of a shape.
In the problem, we have two points: \(P\) at (1,0) and \(Q\) at \((x_1, y_1)\). The midpoint \(M\) is calculated as follows:

• x-coordinate of \(M\) is \(\frac{1 + x_1}{2}\),
• y-coordinate of \(M\) is \(\frac{y_1}{2}\).

This calculation results in the coordinate \(M = \left( \frac{1 + x_1}{2}, \frac{y_1}{2} \right)\), breaking it into a simpler process for students to grasp.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, combines algebra and geometry to describe shapes on a coordinate plane. It enables the use of equations to represent geometric figures. In this branch of mathematics,

• points are plotted using coordinate notation \((x, y)\).

This allows for an algebraic approach to solving geometric problems.
In the context of parabolas and midpoints in this exercise, coordinate geometry helps convert a geometric locus (like a parabola) into an algebraic form \(y^2 = 8x\).
This makes it simpler to work with analytically. With the midpoint formula, coordinate geometry allows us to quickly find a central point \((\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})\) between any two positions in the plane, like points \(P\) and \(Q\). This makes complex geometric relationships simpler and solvable.

Equation of Parabola

A parabola is a unique, symmetric curve that is defined as the set of all points \((x, y)\) that are equidistant from a point called the focus and a line called the directrix. Its equation takes various forms depending on its orientation.

• In this scenario, the equation \(y^2 = 8x\) describes a parabola opening to the right.

When working with parabolic equations like this, the understanding of coordinate relationships becomes important. The equation can be manipulated to express different stories about the curve's path.
In solving the original exercise, we derived the equation for the midpoint's path \(y^2 = 4x - 1\), which represents this new locus as another parabola. The adjustments and simplifications presented in the calculations elucidate how different manipulations can impact the geometric representation of a parabola. Understanding the inherent flexibility of these equations helps in solving many geometry problems.