Equation of the parabola whose vertex is (3, -2) and parallel to x-axis and latusrectum 4 is

Options

A. `(y+2)^(2) = pm 4(x- 1) `
B. `(y+2)^(2) = pm 16(x- 1) `
C. `(y+1)^(2) = pm 4(x- 2) `
D. `(y+1)^(2) = pm 16(x- 2) `

Answer

To find the equation of the parabola whose vertex is (3, -2), is parallel to the x-axis, and has a latus rectum of 4, we can follow these steps:

### Step 1: Identify the form of the equation
Since the parabola is parallel to the x-axis, its equation will be in the form:
\[ (y - y_1)^2 = \pm 4a(x - x_1) \]
where \((x_1, y_1)\) is the vertex of the parabola.

### Step 2: Substitute the vertex coordinates
Given the vertex \((3, -2)\), we can substitute \(x_1 = 3\) and \(y_1 = -2\) into the equation:
\[ (y + 2)^2 = \pm 4a(x - 3) \]

### Step 3: Use the latus rectum to find \(a\)
The latus rectum (L) of the parabola is given as 4. The relationship between the latus rectum and \(a\) is:
\[ L = 4a \]
Substituting \(L = 4\):
\[ 4 = 4a \implies a = 1 \]

### Step 4: Substitute \(a\) back into the equation
Now substituting \(a = 1\) into the equation:
\[ (y + 2)^2 = \pm 4(1)(x - 3) \]
This simplifies to:
\[ (y + 2)^2 = \pm 4(x - 3) \]

### Step 5: Final equation
Thus, the equation of the parabola is:
\[ (y + 2)^2 = 4(x - 3) \quad \text{or} \quad (y + 2)^2 = -4(x - 3) \]
This represents two parabolas, one opening to the right and one opening to the left.