The line `x - b +lambda y = 0` cuts the parabola `y^(2) = 4ax (a gt 0)` at `P(t_(1))` and `Q(t_(2))`. If `b in [2a, 4a]` then range of `t_(1)t_(2)` where `lambda in R` is

Options

A. `[-4,-2]`
B. `[2,4]`
C. `[4,16]`
D. `[-16,-4]`

Explanation

To solve the problem, we need to find the range of the product \( t_1 t_2 \) where the line \( x - b + \lambda y = 0 \) intersects the parabola \( y^2 = 4ax \).

### Step-by-Step Solution:

1. **Write the equations**:
The line can be rewritten as:
\[ x = b - \lambda y \]
The equation of the parabola is:
\[ y^2 = 4ax \]

2. **Substitute the line equation into the parabola**:
Substitute \( x \) from the line equation into the parabola's equation:
\[ y^2 = 4a(b - \lambda y) \]
This simplifies to:
\[ y^2 + 4a\lambda y - 4ab = 0 \]

3. **Identify the coefficients**:
The quadratic equation in \( y \) is:
\[ y^2 + 4a\lambda y - 4ab = 0 \]
Here, the coefficients are:
- \( A = 1 \)
- \( B = 4a\lambda \)
- \( C = -4ab \)

4. **Use Vieta's formulas**:
According to Vieta's formulas, the product of the roots \( t_1 t_2 \) (where \( t_1 \) and \( t_2 \) are the roots corresponding to \( y \)) is given by:
\[ t_1 t_2 = \frac{C}{A} = -4ab \]

5. **Determine the range of \( b \)**:
We know that \( b \) lies in the interval \( [2a, 4a] \). Therefore:
- When \( b = 2a \):
\[ t_1 t_2 = -4a(2a) = -8a^2 \]
- When \( b = 4a \):
\[ t_1 t_2 = -4a(4a) = -16a^2 \]

6. **Establish the range**:
Since \( b \) can take any value in the interval \( [2a, 4a] \), the product \( t_1 t_2 \) will vary from:
\[ -16a^2 \text{ to } -8a^2 \]

Thus, the range of \( t_1 t_2 \) is:
\[ [-16a^2, -8a^2] \]

### Final Answer:
The range of \( t_1 t_2 \) is:
\[ [-16a^2, -8a^2] \]