Normals at P, Q, R are drawn to `y^(2)=4x` which intersect at (3, 0). Then, area of `DeltaPQR`, is

Options

A. `2//5`
B. `1//2`
C. `5//2`
D. 2

Explanation

To solve the problem of finding the area of triangle PQR formed by the normals at points P, Q, and R on the parabola \(y^2 = 4x\) that intersect at the point (3, 0), we will follow these steps:

### Step 1: Understand the parabola and the normals
The given parabola is \(y^2 = 4x\). The parameterization of points on this parabola can be given as:
\[ P(t) = (at^2, 2at) \quad \text{where } a = 1 \text{ (since } 4a = 4 \text{)} \]
Thus, the points on the parabola can be represented as:
\[ P(t) = (t^2, 2t) \]

### Step 2: Write the equation of the normal
The equation of the normal to the parabola at a point \(P(t)\) is:
\[ y + tx = 2a(t + t^3) \]
Substituting \(a = 1\):
\[ y + tx = 2(t + t^3) \]

### Step 3: Substitute the intersection point
Since the normals intersect at the point (3, 0), we substitute \(x = 3\) and \(y = 0\) into the normal equation:
\[ 0 + 3t = 2(t + t^3) \]
This simplifies to:
\[ 3t = 2t + 2t^3 \]
Rearranging gives:
\[ 2t^3 - t = 0 \]
Factoring out \(t\):
\[ t(2t^2 - 1) = 0 \]
Thus, the roots are:
\[ t = 0, \quad t = \frac{1}{\sqrt{2}}, \quad t = -\frac{1}{\sqrt{2}} \]

### Step 4: Find the coordinates of points P, Q, and R
Using the values of \(t\):
1. For \(t_1 = 0\):
\[ P(0) = (0^2, 2 \cdot 0) = (0, 0) \]
2. For \(t_2 = \frac{1}{\sqrt{2}}\):
\[ Q\left(\frac{1}{\sqrt{2}}\right) = \left(\left(\frac{1}{\sqrt{2}}\right)^2, 2 \cdot \frac{1}{\sqrt{2}}\right) = \left(\frac{1}{2}, \sqrt{2}\right) \]
3. For \(t_3 = -\frac{1}{\sqrt{2}}\):
\[ R\left(-\frac{1}{\sqrt{2}}\right) = \left(\left(-\frac{1}{\sqrt{2}}\right)^2, 2 \cdot -\frac{1}{\sqrt{2}}\right) = \left(\frac{1}{2}, -\sqrt{2}\right) \]

### Step 5: Calculate the area of triangle PQR
The vertices of triangle PQR are:
- \(P(0, 0)\)
- \(Q\left(\frac{1}{2}, \sqrt{2}\right)\)
- \(R\left(\frac{1}{2}, -\sqrt{2}\right)\)

Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0\left(\sqrt{2} - (-\sqrt{2})\right) + \frac{1}{2}(-\sqrt{2} - 0) + \frac{1}{2}(0 - \sqrt{2}) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| 0 + \frac{1}{2}(-\sqrt{2}) + \frac{1}{2}(-\sqrt{2}) \right| = \frac{1}{2} \left| -\sqrt{2} \right| = \frac{\sqrt{2}}{2} \]

### Final Area Calculation
The area of triangle PQR is:
\[ \text{Area} = \sqrt{2} \]