Normals of parabola `y^(2)=4x` at P and Q meets at `R(x_(2),0)` and tangents at P and Q meets at `T(x_(1),0)`. If `x_(2)=3`, then find the area of quadrilateral PTQR.

Explanation

To find the area of quadrilateral PTQR formed by the normals and tangents to the parabola \(y^2 = 4x\) at points P and Q, we can follow these steps:

### Step 1: Identify Points P and Q
Let the points \(P\) and \(Q\) on the parabola be represented in terms of parameters \(t_1\) and \(t_2\):
- \(P(t_1^2, 2t_1)\)
- \(Q(t_2^2, 2t_2)\)

### Step 2: Find the Normals at Points P and Q
The equation of the normal to the parabola at point \(P\) is given by:
\[ y = -t_1(x - t_1^2) + 2t_1 \]
Simplifying this gives:
\[ y = -t_1x + 2t_1 + t_1^3 \]
Setting \(y = 0\) to find the x-coordinate where the normal intersects the x-axis:
\[ 0 = -t_1x + 2t_1 + t_1^3 \implies t_1x = 2t_1 + t_1^3 \implies x = 2 + t_1^2 \]

Similarly, for point \(Q\):
\[ y = -t_2(x - t_2^2) + 2t_2 \]
Setting \(y = 0\):
\[ 0 = -t_2x + 2t_2 + t_2^3 \implies t_2x = 2t_2 + t_2^3 \implies x = 2 + t_2^2 \]

### Step 3: Find the Intersection Point R
Given that the normals intersect at \(R(x_2, 0)\) and we know \(x_2 = 3\), we can equate:
\[ 2 + t_1^2 = 3 \implies t_1^2 = 1 \implies t_1 = 1 \text{ or } t_1 = -1 \]
Similarly for \(t_2\):
\[ 2 + t_2^2 = 3 \implies t_2^2 = 1 \implies t_2 = 1 \text{ or } t_2 = -1 \]

### Step 4: Determine Points P and Q
Using \(t_1 = 1\) and \(t_2 = -1\):
- \(P(1^2, 2 \cdot 1) = (1, 2)\)
- \(Q((-1)^2, 2 \cdot (-1)) = (1, -2)\)

### Step 5: Find the Tangents at Points P and Q
The equation of the tangent at point \(P\):
\[ y = t_1(x - t_1^2) + 2t_1 \implies y = 1(x - 1) + 2 = x + 1 \]
Setting \(y = 0\) for the tangent:
\[ 0 = x + 1 \implies x = -1 \]

The equation of the tangent at point \(Q\):
\[ y = -1(x - 1) - 2 \implies y = -x + 1 - 2 = -x - 1 \]
Setting \(y = 0\):
\[ 0 = -x - 1 \implies x = -1 \]

### Step 6: Identify Points T and R
Both tangents meet at point \(T(-1, 0)\) and point \(R(3, 0)\).

### Step 7: Calculate Area of Quadrilateral PTQR
The area of quadrilateral PTQR can be computed as the area of triangles PRT and QRT:
1. Area of triangle PRT:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (1 - (-1)) \times (2 - 0) = \frac{1}{2} \times 2 \times 2 = 2 \]

2. Area of triangle QRT:
\[ \text{Area} = \frac{1}{2} \times (1 - (-1)) \times (-2 - 0) = \frac{1}{2} \times 2 \times 2 = 2 \]

Thus, the total area of quadrilateral PTQR is:
\[ \text{Total Area} = 2 + 2 = 4 \]

### Final Answer
The area of quadrilateral PTQR is \(4\).