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 points \( t_1 \) and \( t_2 \) on the parabola meet the parabola again at point \( t_3 \).

### Step-by-step solution:

1. **Understanding the Parabola**:
The standard equation of a parabola is \( y^2 = 4ax \). The points on the parabola can be represented in parametric form as \( (at^2, 2at) \) where \( t \) is the parameter.

2. **Equation of the Normal**:
The equation of the normal to the parabola at a point \( t \) is given by:
\[ y - 2at = -\frac{1}{a} (x - at^2) \]
Rearranging gives:
\[ y = -\frac{1}{a} x + \left(2a + \frac{at^2}{a}\right) = -\frac{1}{a} x + 2a + at \]

3. **Finding the Normals at Points \( t_1 \) and \( t_2 \)**:
- The normal at \( t_1 \) is:
\[ y = -\frac{1}{a} x + 2a + at_1 \]
- The normal at \( t_2 \) is:
\[ y = -\frac{1}{a} x + 2a + at_2 \]

4. **Finding Intersection Point \( t_3 \)**:
Since both normals meet the parabola again at \( t_3 \), we can set the equations equal to each other:
\[ -\frac{1}{a} x + 2a + at_1 = -\frac{1}{a} x + 2a + at_2 \]
This implies:
\[ at_1 = at_2 \]
Thus, we have \( t_1 = t_2 \) or \( t_1 - t_2 = 0 \).

5. **Equating the Points**:
The coordinates of the intersection point \( t_3 \) can be expressed in terms of \( t_1 \) and \( t_2 \):
\[ t_3 = -t_1 \quad \text{and} \quad t_3 = -t_2 \]
Therefore, we can equate:
\[ -t_1 - \frac{2}{t_1} = -t_2 - \frac{2}{t_2} \]

6. **Cross Multiplying**:
Rearranging gives:
\[ t_1^2 + 2 = t_2^2 + 2 \]
Thus:
\[ t_1^2 = t_2^2 \]

7. **Finding Product \( t_1 t_2 \)**:
From the derived equations, we can see:
\[ t_1^2 t_2 - t_2^2 t_1 = 2(t_1 - t_2) \]
Factoring gives:
\[ (t_1 t_2)(t_1 - t_2) = 2(t_1 - t_2) \]
If \( t_1 \neq t_2 \), we can divide both sides by \( t_1 - t_2 \):
\[ t_1 t_2 = 2 \]

### Conclusion:
Thus, the product \( t_1 t_2 \) equals \( 2 \).