If the line `k^(2)(x-1)+k(y-2)+1=0` touches the parabola `y^(2)-4x-4y+8=0`, then k can be

Options

A. -3
B. `-sqrt(5)`
C. `(7)/(19)`
D. 1000

Explanation

To solve the problem, we need to find the values of \( k \) such that the line \( k^2(x-1) + k(y-2) + 1 = 0 \) touches the parabola given by \( y^2 - 4x - 4y + 8 = 0 \).

### Step 1: Rewrite the equation of the parabola
The first step is to simplify the parabola's equation. We start with:

\[ y^2 - 4y - 4x + 8 = 0 \]

Rearranging gives us:

\[ y^2 - 4y = 4x - 8 \]

Next, we complete the square for the left-hand side:

\[ (y - 2)^2 - 4 = 4x - 8 \]

Adding 4 to both sides results in:

\[ (y - 2)^2 = 4(x - 1) \]

This is the standard form of a parabola that opens to the right with vertex at \( (1, 2) \).

### Step 2: Identify the equation of the tangent line
The equation of the tangent line to the parabola \( (y - k)^2 = 4a(x - h) \) is given by:

\[ y - k = m(x - h) + \frac{a}{m} \]

For our parabola, \( a = 1 \) (since \( 4a = 4 \)), \( h = 1 \), and \( k = 2 \). Thus, the equation of the tangent line becomes:

\[ y - 2 = m(x - 1) + \frac{1}{m} \]

Multiplying through by \( m \) gives:

\[ m(y - 2) = m^2(x - 1) + 1 \]

Rearranging this, we have:

\[ m^2x - my + m + 2 = 0 \]

### Step 3: Compare with the given line equation
Now we compare this with the line equation given in the problem:

\[ k^2(x - 1) + k(y - 2) + 1 = 0 \]

Expanding this gives:

\[ k^2x - k^2 + ky - 2k + 1 = 0 \]

Rearranging leads to:

\[ k^2x + ky - (k^2 - 2k - 1) = 0 \]

### Step 4: Equate coefficients
For the two equations to represent the same line, the coefficients of \( x \) and \( y \) must be equal:

1. Coefficient of \( x \): \( k^2 = m^2 \)
2. Coefficient of \( y \): \( k = -m \)
3. Constant term: \( - (k^2 - 2k - 1) = 0 \)

### Step 5: Solve the equations
From \( k = -m \), we can substitute \( m = -k \) into \( k^2 = m^2 \):

\[ k^2 = (-k)^2 \implies k^2 = k^2 \quad \text{(always true)} \]

Now, substituting \( k = -m \) into the constant term equation:

\[ -(k^2 - 2k - 1) = 0 \implies k^2 - 2k - 1 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:

\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

### Final Values of \( k \)
Thus, the values of \( k \) are:

\[ k = 1 + \sqrt{2} \quad \text{and} \quad k = 1 - \sqrt{2} \]