Problem 76 - A Complete Resource Book in Mathematics for JEE Main 2017

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 is described by the equation \(y^2 - 4x + 2 = 0\). Option (A).

Step 1: Understanding the Locus

The given locus is a parabola described by the equation \(y^2 = 8x\). We need to determine the locus of the midpoint of a segment \(PQ\), where \(P\) is \((1,0)\) and \(Q\) is any point \((x, y)\) on the parabola.

Step 2: Defining the Midpoint

The midpoint \(M\) of the segment joining the points \((1, 0)\) and \((x, y)\) has coordinates \(\left(\frac{1+x}{2}, \frac{0+y}{2}\right)\), which simplifies to \(\left(\frac{1+x}{2}, \frac{y}{2}\right)\).

Step 3: Express \(x\) in Terms of \(y\)

Since point \(Q\) is on the parabola, \(y^2=8x\), we solve for \(x\) in terms of \(y\): \(x = \frac{y^2}{8}\).

Step 4: Substitute \(x\) and Derive Midpoint Equation

Substitute \(x = \frac{y^2}{8}\) into the formula for the x-coordinate of the midpoint: \(\frac{1+x}{2} = \frac{1+\frac{y^2}{8}}{2}\). Simplify to get \(\frac{8+y^2}{16}\). Now, the coordinates of the midpoint \(M\) are \(\left(\frac{8+y^2}{16}, \frac{y}{2}\right)\).

Step 5: Eliminate \(y\) to Find Locus Equation

Set \(\frac{y}{2} = u\), hence \(y = 2u\). Substitute \(y\) in \(\frac{8+y^2}{16} = v\), then substitute \(y=2u\): \(v = \frac{8+(2u)^2}{16} = \frac{8+4u^2}{16}\). Simplifying, we get \(v = \frac{2+u^2}{4}\), which gives the equation \(4v = 2 + u^2\). Rearranging yields \(u^2 - 4v + 2 = 0\).

Step 6: Recognize the Equation of the Locus

Recognize that \(u\) and \(v\) are mappings for \(y\) and \(x\) respectively in the transformed space. This mapping thus transforms back to \(y^2 - 4x + 2 = 0\) in terms of original coordinates of the midpoint.

Key Concepts

Parabola Equation

The parabola is a U-shaped curve that you encounter often in coordinate geometry. In this exercise, the parabola is represented by the equation \(y^2 = 8x\). This equation tells us that it is a sideways opening parabola because the \(y\) term is squared, not \(x\). To visualize, imagine a curve that starts at the origin point, \((0,0)\), and opens out along the x-axis.

Parabolas have a few important properties:

• They are symmetrical about a certain line, known as the axis of symmetry. In our case, since the parabola is sideways, its axis is horizontal.
• The point where the parabola changes direction is called the vertex. Here it's at \((0,0)\).
• The focus (a special point inside the parabola) and the directrix (a line outside the parabola) help define its shape. For equation \(y^2 = 8x\), the focus is \((2, 0)\) and the directrix is \(x = -2\).

This equation comes into play when finding points such as \(Q\) that lie on the parabola, and helps us explore other geometric properties like the locus of midpoints.

Midpoint Calculation

Midpoint calculation is a fundamental concept in coordinate geometry, helping us find the point exactly halfway between two others. For any points \((x_1, y_1)\) and \((x_2, y_2)\), the midpoint is calculated as:

\[\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\]

In this exercise, we have points \((1, 0)\) for \(P\) and \((x, y)\) for \(Q\), where \(Q\) must lie on the parabola. The midpoint \((M)\) thus becomes:

\[\left( \frac{1+x}{2}, \frac{y}{2} \right)\]

This method allows us to deal with geometric problems involving segments and their midpoints easily. It simplifies calculations, guiding us toward the eventual determination of new loci, like in this problem, where we determine the locus of the midpoints.

Coordinate Geometry Locus

The concept of a locus in coordinate geometry refers to the set of all points that satisfy a particular condition or rule. In simpler terms, it's the path traced out by a point that moves according to given rules. In this exercise, we are tasked with finding the locus of the midpoint of segment \(PQ\).

Given that point \(Q\) lies on a parabola \(y^2 = 8x\), you first express \(x\) in terms of \(y\) since \(Q\) must satisfy this parabola equation: \(x = \frac{y^2}{8}\).
Plugging this into the midpoint formula creates a new expression for the set of midpoints \(\left( \frac{8+y^2}{16}, \frac{y}{2} \right)\).

To simplify and represent this set as a locus, we change coordinates. We set \(\frac{y}{2} = u\) and find that \(v = \frac{8+4u^2}{16}\), leading us to the equation \(u^2 - 4v + 2 = 0\).

Finally, transforming \((u, v)\) back to the original \((x, y)\) coordinates, we find that the equation becomes \(y^2 - 4x + 2 = 0\). Therefore, this is the locus we were tasked to find, representing the path of all such midpoints as \(Q\) moves along the parabola.