If the normals drawn at the points `t_(1)` and `t_(2)` on the parabola meet the parabola again at its point `t_(3)`, then `t_(1)t_(2)` equals.

Options

A. 2
B. `-1`
C. `-2`
D. `t_(3) -(2)/(t_(3))`

Answer

To solve the problem, we need to find the product \( t_1 t_2 \) given that the normals drawn at the points \( t_1 \) and \( t_2 \) on the parabola meet the parabola again at the point \( t_3 \).

### Step-by-Step Solution:

1. **Understanding the Normal Equation**:
The equation of the normal to the parabola \( y^2 = 4ax \) at the point corresponding to the parameter \( t \) is given by:
\[ y = -tx + 2a + at^2 \]
For the points \( t_1 \) and \( t_2 \), the normals will intersect the parabola at some point \( t_3 \).

2. **Finding the Intersection Points**:
The point \( t_3 \) can be expressed in terms of \( t_1 \) and \( t_2 \):
\[ t_3 = -\frac{t_1^2 + 2}{t_1} \quad \text{and} \quad t_3 = -\frac{t_2^2 + 2}{t_2} \]
Since both expressions represent the same \( t_3 \), we can set them equal to each other:
\[ -\frac{t_1^2 + 2}{t_1} = -\frac{t_2^2 + 2}{t_2} \]

3. **Cross Multiplying**:
By cross-multiplying, we get:
\[ (t_1^2 + 2)t_2 = (t_2^2 + 2)t_1 \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ t_1^2 t_2 + 2t_2 = t_2^2 t_1 + 2t_1 \]
This can be rearranged to:
\[ t_1^2 t_2 - t_2^2 t_1 = 2t_1 - 2t_2 \]

5. **Factoring the Equation**:
Factoring out common terms, we have:
\[ t_1 t_2 (t_1 - t_2) = 2(t_1 - t_2) \]
If \( t_1 \neq t_2 \), we can divide both sides by \( (t_1 - t_2) \):
\[ t_1 t_2 = 2 \]

6. **Conclusion**:
Therefore, the product \( t_1 t_2 \) equals \( 2 \).

### Final Answer:
\[ t_1 t_2 = 2 \]