If line x-2y-1=0 intersects parabola `y^(2)=4x` at P and Q, then find the point of intersection of normals at P and Q.

Explanation

To solve the problem, we need to find the point of intersection of the normals at points P and Q where the line \( x - 2y - 1 = 0 \) intersects the parabola \( y^2 = 4x \).

### Step 1: Find the points of intersection of the line and the parabola.

The equation of the parabola is given by:
\[ y^2 = 4x \]
The equation of the line can be rearranged to express \( x \) in terms of \( y \):
\[ x = 2y + 1 \]
Now, substitute this expression for \( x \) into the parabola's equation:
\[ y^2 = 4(2y + 1) \]
This simplifies to:
\[ y^2 = 8y + 4 \]
Rearranging gives us:
\[ y^2 - 8y - 4 = 0 \]

### Step 2: Solve the quadratic equation for \( y \).

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -8, c = -4 \):
\[ y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ y = \frac{8 \pm \sqrt{64 + 16}}{2} \]
\[ y = \frac{8 \pm \sqrt{80}}{2} \]
\[ y = \frac{8 \pm 4\sqrt{5}}{2} \]
\[ y = 4 \pm 2\sqrt{5} \]

### Step 3: Find corresponding \( x \) values.

Now, substitute \( y = 4 + 2\sqrt{5} \) and \( y = 4 - 2\sqrt{5} \) back into the line equation to find \( x \):
1. For \( y = 4 + 2\sqrt{5} \):
\[ x = 2(4 + 2\sqrt{5}) + 1 = 8 + 4\sqrt{5} + 1 = 9 + 4\sqrt{5} \]
2. For \( y = 4 - 2\sqrt{5} \):
\[ x = 2(4 - 2\sqrt{5}) + 1 = 8 - 4\sqrt{5} + 1 = 9 - 4\sqrt{5} \]

Thus, the points of intersection \( P \) and \( Q \) are:
\[ P(9 + 4\sqrt{5}, 4 + 2\sqrt{5}), \quad Q(9 - 4\sqrt{5}, 4 - 2\sqrt{5}) \]

### Step 4: Find the normals at points \( P \) and \( Q \).

The slope of the tangent to the parabola \( y^2 = 4x \) at point \( (t^2, 2t) \) is given by:
\[ \frac{dy}{dx} = \frac{2}{y} = \frac{2}{2t} = \frac{1}{t} \]
Thus, the slope of the normal is:
\[ -\frac{1}{\frac{1}{t}} = -t \]

The normal at point \( P(t_1) \) is given by:
\[ y - 2t_1 = -t_1(x - t_1^2) \]
Rearranging gives:
\[ y = -t_1x + t_1^2 + 2t_1 \]

Similarly, for point \( Q(t_2) \):
\[ y = -t_2x + t_2^2 + 2t_2 \]

### Step 5: Find the intersection of the normals.

To find the intersection of the normals, set the two equations equal:
\[ -t_1x + t_1^2 + 2t_1 = -t_2x + t_2^2 + 2t_2 \]
Rearranging gives:
\[ (t_2 - t_1)x = (t_1^2 + 2t_1 - t_2^2 - 2t_2) \]
This can be solved for \( x \) and subsequently for \( y \).

### Step 6: Substitute \( t_1 + t_2 \) and \( t_1 t_2 \).

Using the properties of the roots of the quadratic equation:
- \( t_1 + t_2 = 8 \)
- \( t_1 t_2 = -4 \)

Substituting these values into the equations for the intersection point of the normals gives:
\[ x = 2 + (t_1 + t_2)^2 - t_1 t_2 = 2 + 8^2 - (-4) = 2 + 64 + 4 = 70 \]
\[ y = -t_1 t_2(t_1 + t_2) = -(-4)(8) = 32 \]

Thus, the point of intersection of the normals at points \( P \) and \( Q \) is:
\[ \boxed{(70, 32)} \]