If the image of the point `M(lambda,lambda^(2))` on the line `x+y = lambda^(2)` is N(0, 2), then the sum of the square of all the possible values of `lambda` is equal to

Options

A. 5
B. 2
C. 1
D. 4

Answer

To solve the problem, we need to find the possible values of \(\lambda\) such that the image of the point \(M(\lambda, \lambda^2)\) on the line \(x + y = \lambda^2\) is \(N(0, 2)\). We will use the distance formula from a point to a line to find the values of \(\lambda\).

### Step-by-Step Solution:

1. **Identify the line equation**: The line is given by \(x + y = \lambda^2\). We can rewrite it in the standard form:
\[ x + y - \lambda^2 = 0 \]

2. **Use the distance formula**: The distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 1\), \(B = 1\), \(C = -\lambda^2\), and the point \(N\) is \((0, 2)\).

3. **Calculate the distance from point \(N(0, 2)\) to the line**:
\[ d = \frac{|1 \cdot 0 + 1 \cdot 2 - \lambda^2|}{\sqrt{1^2 + 1^2}} = \frac{|2 - \lambda^2|}{\sqrt{2}} \]

4. **Calculate the coordinates of point \(M(\lambda, \lambda^2)\)**: The distance from \(M\) to the line is:
\[ d' = \frac{|1 \cdot \lambda + 1 \cdot \lambda^2 - \lambda^2|}{\sqrt{2}} = \frac{|\lambda|}{\sqrt{2}} \]

5. **Set the distances equal**: Since \(N\) is the image of \(M\) across the line, the distances must be equal:
\[ \frac{|2 - \lambda^2|}{\sqrt{2}} = \frac{|\lambda|}{\sqrt{2}} \]
We can simplify this to:
\[ |2 - \lambda^2| = |\lambda| \]

6. **Consider two cases for the absolute value**:
- **Case 1**: \(2 - \lambda^2 = \lambda\)
\[ \lambda^2 + \lambda - 2 = 0 \]
Factoring gives:
\[ (\lambda - 1)(\lambda + 2) = 0 \]
Thus, \(\lambda = 1\) or \(\lambda = -2\).

- **Case 2**: \(2 - \lambda^2 = -\lambda\)
\[ \lambda^2 - \lambda - 2 = 0 \]
Factoring gives:
\[ (\lambda - 2)(\lambda + 1) = 0 \]
Thus, \(\lambda = 2\) or \(\lambda = -1\).

7. **Collect all possible values of \(\lambda\)**: The possible values of \(\lambda\) are \(1, -2, 2, -1\).

8. **Calculate the sum of squares**:
\[ 1^2 + (-2)^2 + 2^2 + (-1)^2 = 1 + 4 + 4 + 1 = 10 \]

### Final Answer:
The sum of the square of all the possible values of \(\lambda\) is \(10\).