Consider a parabola `x^2-4xy+4y^2-32x+4y+16=0`.
The focus of the parabola (P) is

Options

A. (2,1)
B. (-2,1)
C. (-2,-1)
D. (2,-1)

Answer

To find the focus of the parabola given by the equation \( x^2 - 4xy + 4y^2 - 32x + 4y + 16 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^2 - 4xy + 4y^2 - 32x + 4y + 16 = 0 \]
We can rearrange this to isolate the quadratic terms on one side:
\[ x^2 - 4xy + 4y^2 = 32x - 4y - 16 \]

### Step 2: Completing the Square
The left-hand side can be recognized as a perfect square:
\[ (x - 2y)^2 = 32x - 4y - 16 \]
This is because \( x^2 - 4xy + 4y^2 = (x - 2y)^2 \).

### Step 3: Further Rearranging
Now, we will add \( 2x - 2y + \lambda^2 \) to both sides to complete the square:
\[ (x - 2y + \lambda)^2 = 32x - 4y - 16 + 2x - 2y + \lambda^2 \]
This gives us:
\[ (x - 2y + \lambda)^2 = (32 + 2\lambda)x + (-4 - 2\lambda)y + (-16 + \lambda^2) \]

### Step 4: Finding the Slopes
Next, we need to find the slopes of the lines formed by the parabola. The slopes of the lines can be derived from the equation:
\[ \text{slope of } l_1 = \frac{1}{2} \]
For the second line, we find:
\[ \text{slope of } l_2 = \frac{32 + 2\lambda}{4 + 4\lambda} \]
Setting the product of the slopes equal to -1 gives us:
\[ \left(\frac{1}{2}\right) \left(\frac{32 + 2\lambda}{4 + 4\lambda}\right) = -1 \]

### Step 5: Solving for Lambda
Cross-multiplying and simplifying:
\[ 32 + 2\lambda = -8 - 8\lambda \]
Bringing all terms involving \(\lambda\) to one side:
\[ 10\lambda = -40 \implies \lambda = -4 \]

### Step 6: Substituting Lambda Back
Substituting \(\lambda = -4\) back into the equations:
\[ x - 2y - 4 = 0 \]
And for the second line:
\[ 32 - 8 = 24 \implies 2x + y = 3 \]

### Step 7: Solving the System of Equations
Now we solve the system of equations:
1. \( x - 2y = 4 \)
2. \( 2x + y = 3 \)

From the first equation, we can express \(x\) in terms of \(y\):
\[ x = 2y + 4 \]
Substituting into the second equation:
\[ 2(2y + 4) + y = 3 \implies 4y + 8 + y = 3 \implies 5y = -5 \implies y = -1 \]
Now substituting back to find \(x\):
\[ x = 2(-1) + 4 = 2 \]

### Step 8: Finding the Focus
Thus, the focus of the parabola is:
\[ (2, -1) \]

### Final Answer
The focus of the parabola \( P \) is \( (2, -1) \).

---