A family of curve S is given by `S -= x^(2) +2xy +y^(2)-4x(1-lambda) - 4y (1+lambda) +4`, then `S = 0` represents (a) pair of straight line `AA lambda in R` (b) straight line for exactly one value of `lambda` (c) parabola `AA lambda in R -{0}` (d) ellipse for three values of `lambda`

Options

A. pair of straight line `AA lambda in R`
B. straight line for exactly one value of `lambda`
C. parabola `AA lambda in R -{0}`
D. ellipse for three values of `lambda`

Explanation

To solve the problem, we need to analyze the equation given by \( S = 0 \) where

\[ S = x^2 + 2xy + y^2 - 4x(1 - \lambda) - 4y(1 + \lambda) + 4. \]

### Step 1: Rewrite the equation
Start by expanding the terms involving \(\lambda\):

\[ S = x^2 + 2xy + y^2 - 4x + 4x\lambda - 4y - 4y\lambda + 4. \]

This simplifies to:

\[ S = x^2 + 2xy + y^2 + 4\lambda x - 4\lambda y - 4x - 4y + 4 = 0. \]

### Step 2: Group terms
Now, we can group the terms:

\[ S = x^2 + 2xy + y^2 + 4\lambda x - 4\lambda y - 4x - 4y + 4 = 0. \]

### Step 3: Complete the square
We can rewrite the quadratic part \(x^2 + 2xy + y^2\) as:

\[ (x + y)^2. \]

Thus, we can express \(S\) as:

\[ S = (x + y)^2 + 4\lambda x - 4\lambda y - 4x - 4y + 4 = 0. \]

### Step 4: Factor the equation
Rearranging gives:

\[ (x + y - 2)^2 + 4\lambda(x - y) = 0. \]

### Step 5: Analyze the conditions
For the equation to hold, both terms must equal zero. This leads to two cases:

1. **Case 1**: If \(\lambda = 0\), then we have:
\[ (x + y - 2)^2 = 0 \implies x + y = 2, \]
which represents a straight line.

2. **Case 2**: If \(\lambda \neq 0\), we can rearrange the equation as:
\[ (x + y - 2)^2 = -4\lambda(x - y). \]
Since the left side is a square, the right side must also be non-negative. This indicates that the equation represents a parabola when \(\lambda \neq 0\).

### Conclusion
From the analysis, we conclude:
- For \(\lambda = 0\), the equation represents a straight line.
- For \(\lambda \neq 0\), the equation represents a parabola.

### Final Answer
Thus, the correct options are:
(b) straight line for exactly one value of \(\lambda\) (when \(\lambda = 0\)).
(c) parabola for all \(\lambda \in \mathbb{R} - \{0\}\).

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