The coordinates of an end-point of the rectum of the parabola `(y-1)^(2)=2(x+2)` are

Options

A. (-2, 1)
B. (-3/2, 1)
C. (-3/2, 2)
D. (-3/2, 0)

Answer

To find the coordinates of an end-point of the rectum of the parabola given by the equation \((y - 1)^2 = 2(x + 2)\), we can follow these steps:

### Step 1: Identify the standard form of the parabola
The given equation can be rewritten in the standard form of a parabola. The standard form for a parabola that opens to the right is given by \((y - k)^2 = 4p(x - h)\), where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus.

### Step 2: Rewrite the equation
The given equation is:
\[ (y - 1)^2 = 2(x + 2) \]
This can be rewritten as:
\[ (y - 1)^2 = 2(x + 2) \implies (y - 1)^2 = 2x + 4 \]
This shows that \(4p = 2\), hence \(p = \frac{1}{2}\).

### Step 3: Identify the vertex
From the equation \((y - 1)^2 = 2(x + 2)\), we can identify the vertex \((h, k)\):
- \(h = -2\)
- \(k = 1\)

Thus, the vertex is at the point \((-2, 1)\).

### Step 4: Find the coordinates of the focus
The focus of the parabola is located at \((h + p, k)\). Since \(p = \frac{1}{2}\):
\[ \text{Focus} = \left(-2 + \frac{1}{2}, 1\right) = \left(-\frac{3}{2}, 1\right) \]

### Step 5: Find the endpoints of the latus rectum
The endpoints of the latus rectum are located at \((h + p, k + 2p)\) and \((h + p, k - 2p)\):
- For the first endpoint:
\[ \left(-\frac{3}{2}, 1 + 2 \cdot \frac{1}{2}\right) = \left(-\frac{3}{2}, 2\right) \]
- For the second endpoint:
\[ \left(-\frac{3}{2}, 1 - 2 \cdot \frac{1}{2}\right) = \left(-\frac{3}{2}, 0\right) \]

### Conclusion
The coordinates of the endpoints of the latus rectum of the parabola \((y - 1)^2 = 2(x + 2)\) are:
\[ \left(-\frac{3}{2}, 2\right) \quad \text{and} \quad \left(-\frac{3}{2}, 0\right) \]