The line `2x+y+lamda=0` is a normal to the parabola `y^(2)=-8x,` is `lamda`=

Options

A. 12
B. -12
C. 24
D. -24

Explanation

To solve the problem, we need to find the value of \(\lambda\) such that the line \(2x + y + \lambda = 0\) is a normal to the parabola given by \(y^2 = -8x\).

### Step-by-Step Solution:

1. **Identify the parameters of the parabola**:
The equation of the parabola is given as \(y^2 = -8x\). We can rewrite this in the standard form \(y^2 = 4ax\) where \(4a = -8\). Thus, we have:
\[ a = -2 \]

2. **Equation of the normal to the parabola**:
The equation of the normal to the parabola \(y^2 = 4ax\) at a point \(P(t)\) is given by:
\[ y + tx = 2a + 2at^2 \]
For our parabola, substituting \(a = -2\):
\[ y + tx = -4 + 2(-2)t^2 \]
Simplifying this gives:
\[ y + tx = -4 - 4t^2 \]
Rearranging, we have:
\[ y = -tx - 4 - 4t^2 \]

3. **Compare with the given line**:
The line is given by:
\[ y = -2x - \lambda \]
We need to equate the two expressions for \(y\):
\[ -tx - 4 - 4t^2 = -2x - \lambda \]

4. **Equate coefficients**:
From the equation, we can equate the coefficients of \(x\) and the constant terms:
- Coefficient of \(x\): \(-t = -2\) implies \(t = 2\).
- Constant terms: \(-4 - 4t^2 = -\lambda\).

5. **Substituting \(t\) into the constant term equation**:
Substitute \(t = 2\) into the constant term equation:
\[ -4 - 4(2^2) = -\lambda \]
Simplifying this gives:
\[ -4 - 16 = -\lambda \]
Thus:
\[ -20 = -\lambda \implies \lambda = 20 \]

6. **Final answer**:
Therefore, the value of \(\lambda\) is:
\[ \lambda = 20 \]