Problem 76 - A Complete Resource Book in Mathematics for JEE Main

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

The locus of the midpoint is \(y^2 - 4x + 2 = 0\), which is option (A).

Step 1: Identify the Equation of the Locus

The given equation of the locus is \(y^2 = 8x\). A point \(Q\) lies on this parabola. Let's denote the coordinates of \(Q\) as \((x_2, y_2)\), where \(y_2^2 = 8x_2\).

Step 2: Determine the Midpoint's Coordinates

Calculate the coordinates of the midpoint \(M\) of segment \(PQ\), where \(P = (1, 0)\) and \(Q = (x_2, y_2)\). The formula for the coordinates of the midpoint \(M\) is given by: \[(x_M, y_M) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).\] Substituting the values, we get: \(x_M = \frac{1 + x_2}{2}\) and \(y_M = \frac{0 + y_2}{2} = \frac{y_2}{2}\).

Step 3: Relate Midpoint Coordinates to Locus Equation

We know \(y_2^2 = 8x_2\). Substituting \(y_2 = 2y_M\), we get \((2y_M)^2 = 8x_2\), which simplifies to \(4y_M^2 = 8x_2\). Divide by 4, yielding \(y_M^2 = 2x_2\).

Step 4: Substitute \(x_2\) in Terms of \(x_M\)

From \(x_M = \frac{1 + x_2}{2}\), we solve for \(x_2\): \(x_2 = 2x_M - 1\). Substitute \(x_2 = 2x_M - 1\) in \(y_M^2 = 2x_2\): \[y_M^2 = 2(2x_M - 1)\] which simplifies to \[y_M^2 = 4x_M - 2.\]

Step 5: Set Locus Equation and Verify Options

Rewrite the equation obtained: \(y_M^2 - 4x_M + 2 = 0\). Compare this expression with the given options. It matches option (A) \(y^2 - 4x + 2 = 0\).

Key Concepts

midpoint coordinate

When you have a line segment with two endpoints and you want to find the middle point, you use the midpoint formula. This formula is helpful when dealing with geometry problems, and it is expressed mathematically as follows: given points \(x_1, y_1\) and \(x_2, y_2\), the midpoint \(M(x_M, y_M)\) is calculated by:

• \(x_M = \frac{x_1 + x_2}{2}\)
• \(y_M = \frac{y_1 + y_2}{2}\)

For the exercise provided, point \(P\) is \( (1, 0) \), and point \(Q\) is on a parabola given by locus \(y^2 = 8x\). When the coordinates of \(Q\) are \( (x_2, y_2) \), the midpoint's coordinates become:

• \(x_M = \frac{1 + x_2}{2}\)
• \(y_M = \frac{y_2}{2}\)

These values are crucial to find out how the midpoint moves when point \(Q\) changes along the curve.

parabola locus

A parabola is a U-shaped curve that has various important properties, one of them being its equation. In coordinate geometry, the standard equation of a parabola opening to the right is \(y^2 = 4ax\). The parabola described in the exercise follows a similar equation: \(y^2 = 8x\). From this, it's apparent that the parabola opens to the right, and the term 8 represents \(4a\), meaning \(a = 2\). The point \(Q(x_2, y_2)\) lies on this parabola, satisfying the equation \(y_2^2 = 8x_2\). This connection is needed to explore how the coordinates of the midpoint \(M\) relate to the path traced by \(Q\).

• The parabola's movement can be defined by how \(y_2\) and \(x_2\) satisfy the equation.
• By introducing the midpoint, we can reflect its symmetry and calculate its own path-locus equation based on changes in \(Q\).

Understanding the parabola's locus is key to navigating problems involving midpoint displacement in JEE Main Mathematics.

JEE Main mathematics

The Joint Entrance Examination (JEE Main) is one of India's most prestigious engineering college entrance exams. Questions in this exam, such as the one you've been provided, assess understanding of mathematical concepts like geometry, algebra, and their intersections. Problems often include finding equations of loci, solving for midpoints, and understanding the behavior of curves like parabolas. Some essential factors while preparing for a topic like the locus equation of a parabola in JEE Main include:

• Grasp of coordinate geometry fundamentals, including point and line behavior.
• Ability to manipulate algebraic expressions for complex result derivations, as seen when relating midpoint coordinates.
• Understanding real-world applications of these shapes and their path—a trait necessary to tackle different versions of these problems.

For concepts like midpoint coordinate or parabola locus, being able to seamlessly connect algebraic and geometric interpretations is pivotal.