If the chord joining the points `t_1`and `t_2` on the parabola `y^2 = 4ax` subtends a right angle at its vertex then `t_1t_2=`

Options

A. 0
B. -4
C. -1
D. 2

Explanation

To solve the problem, we need to find the value of \( t_1 t_2 \) given that the chord joining the points \( t_1 \) and \( t_2 \) on the parabola \( y^2 = 4ax \) subtends a right angle at the vertex (origin).

### Step-by-Step Solution:

1. **Understand the Parametric Form of the Parabola**:
The points on the parabola \( y^2 = 4ax \) can be expressed in parametric form as:
\[ P(t_1) = (at_1^2, 2at_1) \quad \text{and} \quad Q(t_2) = (at_2^2, 2at_2) \]

2. **Find the Slopes of the Lines OP and OQ**:
The slope of the line OP (from the origin to point P) is given by:
\[ m_{OP} = \frac{2at_1 - 0}{at_1^2 - 0} = \frac{2t_1}{t_1^2} \]
The slope of the line OQ (from the origin to point Q) is given by:
\[ m_{OQ} = \frac{2at_2 - 0}{at_2^2 - 0} = \frac{2t_2}{t_2^2} \]

3. **Use the Condition for Right Angles**:
Since the chord subtends a right angle at the vertex, the product of the slopes must equal -1:
\[ m_{OP} \cdot m_{OQ} = -1 \]
Substituting the slopes we found:
\[ \left(\frac{2t_1}{t_1^2}\right) \cdot \left(\frac{2t_2}{t_2^2}\right) = -1 \]

4. **Simplify the Equation**:
This simplifies to:
\[ \frac{4t_1t_2}{t_1^2t_2^2} = -1 \]
Rearranging gives:
\[ 4t_1t_2 = -t_1^2t_2^2 \]

5. **Rearranging the Equation**:
We can rearrange this to:
\[ t_1^2t_2^2 + 4t_1t_2 = 0 \]
Factoring out \( t_1t_2 \):
\[ t_1t_2(t_1t_2 + 4) = 0 \]

6. **Finding the Values**:
This gives us two cases:
- \( t_1t_2 = 0 \) (not applicable since both points are on the parabola)
- \( t_1t_2 + 4 = 0 \) which leads to:
\[ t_1t_2 = -4 \]

### Final Answer:
Thus, the value of \( t_1 t_2 \) is:
\[ \boxed{-4} \]