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

Options

A. `4a^2`
B. `2a^2`
C. `-4a^2`
D. `8a^2`

Answer

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

### Step-by-Step Solution:

1. **Identify the Points on the Parabola:**
Let the points P and Q on the parabola be represented by their parameters \(t_1\) and \(t_2\). The coordinates of these points can be expressed as:
- \(P(t_1) = (at_1^2, 2at_1)\)
- \(Q(t_2) = (at_2^2, 2at_2)\)

2. **Equation of the Normal at Point P:**
The equation of the normal to the parabola at point P is given by:
\[ y - 2at_1 = -\frac{2a}{2a} (x - at_1^2) \]
Simplifying this gives:
\[ y - 2at_1 = -x + at_1^2 \]
Rearranging, we get:
\[ x + y = at_1^2 + 2at_1 \quad \text{(Equation 1)} \]

3. **Equation of the Normal at Point Q:**
Similarly, the equation of the normal at point Q is:
\[ y - 2at_2 = -\frac{2a}{2a} (x - at_2^2) \]
Simplifying gives:
\[ y - 2at_2 = -x + at_2^2 \]
Rearranging, we get:
\[ x + y = at_2^2 + 2at_2 \quad \text{(Equation 2)} \]

4. **Intersection Point R:**
Since the normals at P and Q intersect at point R, we can set the right-hand sides of Equations 1 and 2 equal to each other:
\[ at_1^2 + 2at_1 = at_2^2 + 2at_2 \]
Rearranging gives:
\[ at_1^2 - at_2^2 + 2a(t_1 - t_2) = 0 \]
Factoring out \(a\):
\[ a(t_1^2 - t_2^2 + 2(t_1 - t_2)) = 0 \]
Since \(a \neq 0\), we have:
\[ t_1^2 - t_2^2 + 2(t_1 - t_2) = 0 \]
This can be factored as:
\[ (t_1 - t_2)(t_1 + t_2 + 2) = 0 \]
Thus, either \(t_1 = t_2\) (not possible as P and Q are distinct points) or:
\[ t_1 + t_2 + 2 = 0 \implies t_1 + t_2 = -2 \]

5. **Product of the Ordinates:**
The ordinates of points P and Q are given by:
- Ordinate of P: \(y_1 = 2at_1\)
- Ordinate of Q: \(y_2 = 2at_2\)

Therefore, the product of the ordinates is:
\[ y_1 \cdot y_2 = (2at_1)(2at_2) = 4a^2(t_1 t_2) \]

6. **Finding \(t_1 t_2\):**
From the equation \(t_1 + t_2 = -2\), we can use the identity:
\[ (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2 \]
Substituting \(t_1 + t_2 = -2\):
\[ (-2)^2 = t_1^2 + t_2^2 + 2t_1t_2 \]
Thus:
\[ 4 = t_1^2 + t_2^2 + 2t_1t_2 \]
From the earlier factored equation, we can find \(t_1 t_2\):
\[ t_1 t_2 = -2 \]

7. **Final Calculation:**
Therefore:
\[ y_1 \cdot y_2 = 4a^2(-2) = -8a^2 \]

### Conclusion:
The product of the ordinates of points P and Q is \(-8a^2\).