The locus of the intersection points of pair of perpendicular tangents to the parabola `x^2-4x + 4y-8=0`. is

Options

A. `y-4=0`
B. `x-3=0`
C. `x+y=2`
D. None of these

Answer

To find the locus of the intersection points of pairs of perpendicular tangents to the parabola given by the equation \(x^2 - 4x + 4y - 8 = 0\), we will follow these steps:

### Step 1: Rewrite the equation of the parabola
We start with the equation:
\[ x^2 - 4x + 4y - 8 = 0 \]
We can rearrange this to isolate \(y\):
\[ 4y = -x^2 + 4x + 8 \]
\[ y = -\frac{1}{4}x^2 + x + 2 \]
This is a downward-opening parabola.

### Step 2: Find the vertex of the parabola
The vertex form of a parabola is given by the formula \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. We can complete the square for the quadratic in \(x\):
\[ y = -\frac{1}{4}(x^2 - 4x) + 2 \]
Completing the square inside the parentheses:
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Substituting back:
\[ y = -\frac{1}{4}((x - 2)^2 - 4) + 2 \]
\[ y = -\frac{1}{4}(x - 2)^2 + 1 + 2 \]
\[ y = -\frac{1}{4}(x - 2)^2 + 3 \]
Thus, the vertex of the parabola is at \((2, 3)\).

### Step 3: Determine the equation of the tangents
The general equation of the tangent to the parabola \(y = ax^2 + bx + c\) at point \((x_1, y_1)\) is given by:
\[ y - y_1 = m(x - x_1) \]
For perpendicular tangents, the slopes \(m_1\) and \(m_2\) of the tangents satisfy:
\[ m_1 \cdot m_2 = -1 \]

### Step 4: Find the locus of intersection points
To find the locus of the intersection points of the perpendicular tangents, we can use the property that the locus of the intersection points of the tangents to a parabola is another parabola.

Given that the vertex of the original parabola is at \((2, 3)\), we can derive the locus of the intersection points of the perpendicular tangents, which will be a horizontal line due to symmetry.

### Step 5: Finalize the locus equation
From the previous steps, we can conclude that the locus of the intersection points of the pair of perpendicular tangents to the parabola is:
\[ y = 4 \]
This can be written as:
\[ y - 4 = 0 \]

### Conclusion
Thus, the locus of the intersection points of the pair of perpendicular tangents to the parabola \(x^2 - 4x + 4y - 8 = 0\) is:
\[ \boxed{y - 4 = 0} \]