If the normals at two points P and Q of a parabola `y^2 = 4x` intersect at a third point R on the parabola `y^2 = 4x`, then the product of the ordinates of P and Q is equal to

Answer

To solve the problem, we need to find the product of the ordinates of points P and Q on the parabola \( y^2 = 4x \) where the normals at these points intersect at a third point R on the parabola.

### Step-by-Step Solution:

1. **Identify the points on the parabola**:
Let the points P and Q on the parabola \( y^2 = 4x \) be represented by their parameters \( t_1 \) and \( t_2 \). The coordinates of these points can be expressed as:
\[ P(t_1) = (t_1^2, 2t_1) \quad \text{and} \quad Q(t_2) = (t_2^2, 2t_2) \]

2. **Equation of the normals**:
The equation of the normal to the parabola at point \( (t^2, 2t) \) is given by:
\[ y = -tx + 2t + t^2 \]
Therefore, the equations of the normals at points P and Q are:
\[ y = -t_1x + 2t_1 + t_1^2 \quad \text{(for point P)} \]
\[ y = -t_2x + 2t_2 + t_2^2 \quad \text{(for point Q)} \]

3. **Finding the intersection point R**:
To find the intersection point R of the two normals, we can set the equations equal to each other:
\[ -t_1x + 2t_1 + t_1^2 = -t_2x + 2t_2 + t_2^2 \]
Rearranging gives:
\[ (t_2 - t_1)x = 2(t_2 - t_1) + (t_2^2 - t_1^2) \]
Factoring the right side:
\[ (t_2 - t_1)x = (t_2 - t_1)(2 + t_2 + t_1) \]
If \( t_2 \neq t_1 \), we can divide both sides by \( t_2 - t_1 \):
\[ x = 2 + t_1 + t_2 \]

4. **Finding the y-coordinate of R**:
Substitute \( x \) back into one of the normal equations to find the y-coordinate:
\[ y = -t_1(2 + t_1 + t_2) + 2t_1 + t_1^2 \]
Simplifying this gives:
\[ y = -2t_1 - t_1^2 - t_1t_2 + 2t_1 + t_1^2 = -t_1t_2 \]

5. **Determine the product of the ordinates**:
The ordinates of points P and Q are \( 2t_1 \) and \( 2t_2 \) respectively. Therefore, the product of the ordinates is:
\[ (2t_1)(2t_2) = 4t_1t_2 \]

6. **Using the properties of the roots**:
From the earlier steps, we know that the product of the roots \( t_1 \) and \( t_2 \) of the quadratic equation formed by the normals is equal to 2. Hence:
\[ t_1t_2 = 2 \]
Therefore, substituting this back gives:
\[ 4t_1t_2 = 4 \times 2 = 8 \]

### Final Answer:
The product of the ordinates of points P and Q is \( 8 \).