The number of distinct normals that be drawn from (-2,1) to the parabola `y^(2)-4x-2y-3=0` is:

Options

A. 1
B. 2
C. 3
D. 0

Explanation

To solve the problem of finding the number of distinct normals that can be drawn from the point (-2, 1) to the parabola given by the equation \( y^2 - 4x - 2y - 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the parabola in standard form
We start with the equation of the parabola:
\[ y^2 - 4x - 2y - 3 = 0 \]

Rearranging the equation, we can group the \( y \) terms:
\[ y^2 - 2y = 4x + 3 \]

Next, we complete the square for the \( y \) terms:
\[ (y^2 - 2y + 1) = 4x + 3 + 1 \]
\[ (y - 1)^2 = 4x + 4 \]
\[ (y - 1)^2 = 4(x + 1) \]

This is now in the standard form of a parabola:
\[ (y - k)^2 = 4a(x - h) \]
where the vertex is at \( (-1, 1) \).

### Step 2: Identify the vertex and axis of the parabola
From the standard form, we can identify:
- Vertex: \( (-1, 1) \)
- Axis of symmetry: \( y = 1 \)

### Step 3: Analyze the point (-2, 1)
The point from which we want to draw the normals is \( (-2, 1) \). We observe that this point lies on the line \( y = 1 \), which is also the axis of symmetry of the parabola.

### Step 4: Determine the number of normals
Since the point \( (-2, 1) \) lies on the axis of the parabola, we can only draw one normal line from this point to the parabola. This is because the normal at the vertex of the parabola is vertical, and since our point lies directly on the axis, it can only intersect the parabola at one point.

### Conclusion
Thus, the number of distinct normals that can be drawn from the point \( (-2, 1) \) to the parabola is:
**1**