A parabola is drawn with focus at (3,4) and vertex at the focus of the parabola `y^2-12x-4y+4=0`. The equation of the parabola is

Options

A. `x^2-6x+8y+25=0`
B. `y^2-8x-6y+25=0`
C. `x^2-6x+8y-25=0`
D. `x^2+6x-8y-25=0`

Answer

To find the equation of the parabola with a focus at (3, 4) and a vertex at the focus of the parabola given by the equation \( y^2 - 12x - 4y + 4 = 0 \), we can follow these steps:

### Step 1: Rewrite the given parabola equation
Start with the equation of the parabola:
\[ y^2 - 12x - 4y + 4 = 0 \]
Rearranging it gives:
\[ y^2 - 4y + 4 = 12x \]

### Step 2: Complete the square for the y terms
To complete the square for the left side:
\[ (y - 2)^2 = 12x \]

### Step 3: Identify the vertex and focus of the given parabola
From the equation \((y - 2)^2 = 12x\), we can identify:
- The vertex of this parabola is at \((0, 2)\).
- The focus can be calculated using the formula for the focus of a parabola \( (h + p, k) \), where \( p \) is the distance from the vertex to the focus. Here, \( p = 3 \) (since \( 12 = 4p \) implies \( p = 3 \)). Thus, the focus is:
\[ (0 + 3, 2) = (3, 2) \]

### Step 4: Determine the vertex of the required parabola
The vertex of the required parabola is given to be at the focus of the parabola we just analyzed, which is at \((3, 2)\).

### Step 5: Write the equation of the required parabola
Now, we have:
- Focus: \((3, 4)\)
- Vertex: \((3, 2)\)

Since the focus is above the vertex, the parabola opens upwards. The standard form of a parabola that opens upwards is:
\[ (x - h)^2 = 4p(y - k) \]
where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus.

Here:
- \(h = 3\)
- \(k = 2\)
- \(p = 4 - 2 = 2\)

Thus, substituting these values into the equation gives:
\[ (x - 3)^2 = 4 \cdot 2 (y - 2) \]
This simplifies to:
\[ (x - 3)^2 = 8(y - 2) \]

### Final Equation
The equation of the required parabola is:
\[ (x - 3)^2 = 8(y - 2) \]