If the normals at the end points of a variable chord PQ of the parabola `y^(2)-4y-2x=0` are perpendicular, then the tangents at P and Q will intersect at

Options

A. x + y = 3
B. 3x - 7 = 0
C. y + 3 = 0
D. 2x + 5 = 0

Answer

To solve the problem, we need to analyze the given parabola and the conditions of the normals and tangents at points P and Q.

### Step-by-Step Solution:

1. **Rearranging the Parabola Equation:**
The given equation of the parabola is:
\[ y^2 - 4y - 2x = 0 \]
Rearranging it, we get:
\[ y^2 - 4y = 2x \implies (y - 2)^2 = 2(x + 2) \]
This is in the standard form of a parabola \((y - k)^2 = 4a(x - h)\) where \(a = \frac{1}{2}\) and the vertex is at \((-2, 2)\).

2. **Finding Points P and Q:**
Let the points P and Q on the parabola be represented in terms of parameters \(t_1\) and \(t_2\):
\[ P = (-2 + \frac{1}{2}t_1^2, 2 + t_1) \]
\[ Q = (-2 + \frac{1}{2}t_2^2, 2 + t_2) \]

3. **Finding the Slopes of the Normals:**
The slope of the normal at point P is:
\[ m_1 = -t_1 \]
The slope of the normal at point Q is:
\[ m_2 = -t_2 \]
Since the normals are perpendicular, we have:
\[ m_1 \cdot m_2 = -1 \implies (-t_1)(-t_2) = -1 \implies t_1 t_2 = -1 \]

4. **Finding the Tangents at P and Q:**
The equations of the tangents at points P and Q can be expressed as:
\[ t_1(y - 2) = x + 2 - \frac{1}{2}t_1^2 \]
\[ t_2(y - 2) = x + 2 - \frac{1}{2}t_2^2 \]

5. **Setting Up the Tangent Equations:**
Rearranging the tangent equations:
\[ t_1y - 2t_1 = x + 2 - \frac{1}{2}t_1^2 \implies x - t_1y + (2t_1 + 2 - \frac{1}{2}t_1^2) = 0 \]
\[ t_2y - 2t_2 = x + 2 - \frac{1}{2}t_2^2 \implies x - t_2y + (2t_2 + 2 - \frac{1}{2}t_2^2) = 0 \]

6. **Finding the Intersection of the Tangents:**
To find the intersection of the tangents, we can solve the two equations simultaneously. The intersection point \((x, y)\) can be found by eliminating \(x\) from the two tangent equations.

7. **Finding the Coordinates of Intersection:**
After solving, we find that the x-coordinate of the intersection point is:
\[ h = -2 - a \cdot t_1 t_2 \]
Given \(t_1 t_2 = -1\) and \(a = \frac{1}{2}\):
\[ h = -2 - \frac{1}{2}(-1) = -2 + \frac{1}{2} = -\frac{5}{2} \]
Therefore, the locus of the intersection is:
\[ 2x + 5 = 0 \]

### Final Answer:
The tangents at P and Q will intersect at the locus given by:
\[ 2x + 5 = 0 \]