The equation of parabola whose vertex and focus are `(0,4) and (0,2)` respectively, is

Options

A. `y ^(2) + 8x =32`
B. `y ^(2) - 8x =32`
C. `x ^(2) + 8y =32`
D. `x ^(2) - 8y =32`

Answer

To find the equation of the parabola with the given vertex and focus, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Vertex and Focus**:
- The vertex of the parabola is given as \( V(0, 4) \).
- The focus of the parabola is given as \( F(0, 2) \).

2. **Determine the Orientation of the Parabola**:
- Since the vertex is above the focus (the y-coordinate of the vertex is greater than that of the focus), the parabola opens downwards.

3. **Find the Directrix**:
- The distance between the vertex and the focus is \( |4 - 2| = 2 \).
- Since the vertex is above the focus, the directrix will be located at a distance of 2 units above the vertex. Thus, the equation of the directrix is:
\[ y = 4 + 2 = 6 \]

4. **Use the Standard Form of the Parabola**:
- The standard form of a parabola that opens downwards is given by:
\[ (x - h)^2 = -4p(y - k) \]
where \( (h, k) \) is the vertex and \( p \) is the distance from the vertex to the focus (which is 2 in this case).

5. **Substitute the Values**:
- Here, \( h = 0 \), \( k = 4 \), and \( p = 2 \). Thus, we have:
\[ (x - 0)^2 = -4(2)(y - 4) \]
Simplifying this gives:
\[ x^2 = -8(y - 4) \]

6. **Rearranging the Equation**:
- Expanding the equation:
\[ x^2 = -8y + 32 \]
- Rearranging gives:
\[ x^2 + 8y = 32 \]

### Final Equation:
The equation of the parabola is:
\[ x^2 + 8y = 32 \]