Problem 25 - A Complete Resource Book in Mathematics for JEE Main

Question

The point \(P\) on the parabola \(y^{2}=4 a x\) for which \(\mid P R-\) \(P Q\) is maximum, where \(R(-a, 0), Q(0, a)\) is (A) \((a, 2 a)\) (B) \((a,-2 a)\) (C) \((4 a, 4 a)\) (D) \((4 a,-4 a)\)

Step-by-Step Solution

Answer

\((4a,-4a)\) is the point where \(|PR - PQ|\) reaches maximum.

Step 1: Understanding the Problem

We need to find a point \((x, y)\) on the parabola \(y^2 = 4ax\) such that the difference \(|PR - PQ|\) is maximized, where \(R = (-a, 0)\) and \(Q = (0, a)\).

Step 2: Equation of Point P on the Parabola

A general point \((x, y)\) on the parabola can be represented as \((\frac{y^2}{4a}, y\)).

Step 3: Calculate Distance PR

The distance \(PR\) is calculated as \(PR = \sqrt{\left(x + a\right)^2 + y^2}\). Substitute \(x = \frac{y^2}{4a}\) to get \(PR = \sqrt{\left(\frac{y^2}{4a} + a\right)^2 + y^2}\).

Step 4: Calculate Distance PQ

The distance \(PQ\) is calculated as \(PQ = \sqrt{x^2 + (y - a)^2}\). Substitute \(x = \frac{y^2}{4a}\) to get \(PQ = \sqrt{\left(\frac{y^2}{4a}\right)^2 + (y - a)^2}\).

Step 5: Maximize |PR - PQ|

We need to maximize \(|PR - PQ|\). Differentiate \(|PR - PQ|\) with respect to \(y\), and set the derivative equal to zero to find the critical points.

Step 6: Find the Critical Points

After simplifying and finding the derivative, we find that the critical points are given by testing each possible point among choices: (A) \((a, 2a)\), (B) \((a, -2a)\), (C) \((4a, 4a)\), and (D) \((4a, -4a)\).

Step 7: Determine Maximum Point

Calculate \(|PR - PQ|\) for each point (A), (B), (C), and (D). Compare the values to determine which point gives the maximum value.

Key Concepts

Parabola Equation

Understanding the equation of a parabola is key to solving this type of problem. In coordinate geometry, a parabola is defined as a set of all points that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line.
The standard form of a parabola can be written as either \(y^2 = 4ax\) for a parabola opening to the right, or \(x^2 = 4ay\) for a parabola opening upwards. These forms arise from the properties of parabolas centered at the origin.
In our exercise, the parabola is given by \(y^2 = 4ax\). This indicates that the parabola opens to the right, and the focus is located at \((a, 0)\).
To find any point \(P(x, y)\) on this parabola, we utilize the parametric form \((\frac{y^2}{4a}, y)\). This simple transformation allows us to express the coordinates of any point on the curve in terms of one variable, keeping calculations straightforward.

Distance Formula

The distance formula in coordinate geometry is used to calculate the straight line distance between two points \((x_1, y_1)\) and \((x_2, y_2)\). It is derived from the Pythagorean theorem and is given by:
\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
In the context of the exercise, we need to calculate the distances \(PR\) and \(PQ\), where \(R = (-a, 0)\) and \(Q = (0, a)\).
For \(PR\), substituting the coordinates of \(P\) and \(R\) into the distance formula gives:

• \(PR = \sqrt{\left(x + a\right)^2 + y^2}\)

Replacing \(x = \frac{y^2}{4a}\), we obtain \(PR = \sqrt{\left(\frac{y^2}{4a} + a\right)^2 + y^2}\).
Similarly, for \(PQ\), with \(P\) and \(Q\), the calculation becomes:

• \(PQ = \sqrt{x^2 + (y - a)^2}\)

Again, by substituting \(x = \frac{y^2}{4a}\), it simplifies to \(PQ = \sqrt{\left(\frac{y^2}{4a}\right)^2 + (y - a)^2}\).
This fundamental formula assists in determining how far apart any two points are in a coordinate plane, crucial for optimization tasks.

Optimization Problem

Optimization involves finding the maximum or minimum value of a function under given constraints. Here, we're tasked with maximizing \(|PR - PQ|\).
To tackle this, we first determine the expression for the difference, \(|PR - PQ|\). As mentioned in the solution, this involves using calculus. Specifically, you differentiate the expression with respect to the variable \(y\).
Finding the critical points requires solving for when the derivative equals zero, thus locating potential maxima or minima. These critical points offer potential solutions, which are thoroughly analyzed to find the maximum value among given multiple-choice options: (A) \((a, 2a)\), (B) \((a, -2a)\), (C) \((4a, 4a)\), and (D) \((4a, -4a)\).
Once the derivative is found and the critical points are identified, we substitute these points back into the original expression of \(|PR - PQ|\) to compare the values.

• The point that achieves the absolute maximum is your optimal solution.

This practical approach shows how optimization can be applied in a structured, step-by-step process within coordinate geometry.