Consider two points `A(at_1^2,2at_1) and B(at_2^2,2at_2)` lying on the parabola `y^2 = 4ax.` If the line joining the points `A and B` passes through the point `P(b,o).` then `t_1 t_2` is equal to :

Options

A. `(a)/(b)`
B. `-(a)/(b)`
C. `(b)/(a)`
D. `-(b)/(a)`

Answer

To solve the problem, we need to find the product \( t_1 t_2 \) given two points \( A(at_1^2, 2at_1) \) and \( B(at_2^2, 2at_2) \) on the parabola \( y^2 = 4ax \) and the condition that the line joining these points passes through the point \( P(b, 0) \).

### Step-by-Step Solution:

1. **Identify the Points**:
The points given are:
- \( A(at_1^2, 2at_1) \)
- \( B(at_2^2, 2at_2) \)

2. **Equation of the Line**:
The equation of the line joining points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) can be expressed as:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting the coordinates of points \( A \) and \( B \):
\[ y - 2at_1 = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2}(x - at_1^2) \]

3. **Simplify the Equation**:
The slope \( \frac{2a(t_2 - t_1)}{a(t_2^2 - t_1^2)} \) simplifies to:
\[ \frac{2(t_2 - t_1)}{t_2^2 - t_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_2 + t_1} \]
Thus, the equation becomes:
\[ y - 2at_1 = \frac{2}{t_2 + t_1}(x - at_1^2) \]

4. **Cross Multiply**:
Rearranging gives:
\[ (t_2 + t_1)(y - 2at_1) = 2(x - at_1^2) \]
Expanding this:
\[ (t_2 + t_1)y - 2at_1(t_2 + t_1) = 2x - 2at_1^2 \]

5. **Substituting Point \( P(b, 0) \)**:
Since the line passes through \( P(b, 0) \), substitute \( x = b \) and \( y = 0 \):
\[ (t_2 + t_1)(0 - 2at_1) = 2b - 2at_1^2 \]
This simplifies to:
\[ -2at_1(t_2 + t_1) = 2b - 2at_1^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2b = 2at_1(t_2 + t_1) + 2at_1^2 \]
Dividing by 2:
\[ b = at_1(t_2 + t_1 + t_1) = at_1(t_2 + 2t_1) \]

7. **Finding \( t_1 t_2 \)**:
From the equation \( b = at_1(t_2 + 2t_1) \), we can isolate \( t_1 t_2 \):
\[ t_1 t_2 = -\frac{b}{a} \]

### Final Answer:
Thus, the product \( t_1 t_2 \) is:
\[ t_1 t_2 = -\frac{b}{a} \]