To find the area of the quadrilateral PTQR formed by the normals and tangents of the parabola \(y^2 = 4x\), we can follow these steps:
### Step 1: Identify Points P and Q on the Parabola
The parabola given is \(y^2 = 4x\). The general points on the parabola can be represented as:
- Point \(P(t_1) = (at_1^2, 2at_1)\)
- Point \(Q(t_2) = (at_2^2, 2at_2)\)
Since \(a = 1\) for the parabola \(y^2 = 4x\), we have:
- \(P(t_1) = (t_1^2, 2t_1)\)
- \(Q(t_2) = (t_2^2, 2t_2)\)
### Step 2: Find the Equation of the Normals at Points P and Q
The equation of the normal at point \(P(t_1)\) is given by:
\[
y = -t_1x + 2t_1 + t_1^3
\]
Similarly, the equation of the normal at point \(Q(t_2)\) is:
\[
y = -t_2x + 2t_2 + t_2^3
\]
### Step 3: Find the Intersection Point R of Normals
The normals meet at point \(R(x_2, 0)\). Given that \(x_2 = 3\), we can substitute \(x = 3\) into the equations of the normals to find \(t_1\) and \(t_2\).
Substituting \(x = 3\) into the normal at \(P\):
\[
0 = -t_1(3) + 2t_1 + t_1^3
\]
This simplifies to:
\[
t_1^3 - t_1 = 3t_1 \quad \Rightarrow \quad t_1^3 - 4t_1 = 0 \quad \Rightarrow \quad t_1(t_1^2 - 4) = 0
\]
Thus, \(t_1 = 0, 2, -2\).
### Step 4: Find Points P and Q
Assuming \(t_1 = 2\) and \(t_2 = -2\) (the other values can be checked similarly):
- \(P(2) = (2^2, 2 \cdot 2) = (4, 4)\)
- \(Q(-2) = ((-2)^2, 2 \cdot -2) = (4, -4)\)
### Step 5: Find the Intersection Point T of Tangents
The tangent at point \(P\) is:
\[
y = 2(x - 4) + 4 \quad \Rightarrow \quad y = 2x - 4
\]
The tangent at point \(Q\) is:
\[
y = -2(x - 4) - 4 \quad \Rightarrow \quad y = -2x + 4
\]
Setting these equal to find \(T(x_1, 0)\):
\[
2x - 4 = -2x + 4
\]
Solving for \(x\):
\[
4x = 8 \quad \Rightarrow \quad x = 2
\]
Thus, \(T(2, 0)\).
### Step 6: Calculate Area of Quadrilateral PTQR
The area of quadrilateral PTQR can be calculated as the sum of the areas of triangles PTR and QTR.
1. **Area of Triangle PTR**:
- Base \(PT = 4 - 2 = 2\)
- Height from \(R(3, 0)\) to line \(PT\) (which is horizontal) is 4.
- Area = \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 4 = 4\).
2. **Area of Triangle QTR**:
- Base \(QT = 4 - 2 = 2\)
- Height from \(R(3, 0)\) to line \(QT\) (which is horizontal) is 4.
- Area = \(\frac{1}{2} \times base \times height = \frac{1}{2} \times 2 \times 4 = 4\).
### Final Area Calculation
Total Area of Quadrilateral PTQR:
\[
\text{Area} = \text{Area of } PTR + \text{Area of } QTR = 4 + 4 = 8
\]
Thus, the area of quadrilateral PTQR is \(8\) square units.
