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 the end-points 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 \((y-1)^{2} = 2(x+2)\) can be rewritten in the standard form of a parabola that opens to the right, which is \((y-k)^{2} = 4p(x-h)\). Here, \((h, k)\) is the vertex of the parabola.

### Step 2: Determine the vertex
From the equation \((y-1)^{2} = 2(x+2)\):
- We can see that \(h = -2\) and \(k = 1\).
- Therefore, the vertex of the parabola is at the point \((-2, 1)\).

### Step 3: Find the value of \(p\)
The equation can be compared to the standard form:
- Here, \(4p = 2\), which gives us \(p = \frac{1}{2}\).

### Step 4: Identify the coordinates of the end-points of the rectum
For a parabola that opens to the right, the end-points of the rectum are located at:
- \( (h + p, k + 2p) \) and \( (h + p, k - 2p) \).

### Step 5: Calculate the coordinates
Substituting the values of \(h\), \(k\), and \(p\):
1. For the first end-point:
\[ (h + p, k + 2p) = \left(-2 + \frac{1}{2}, 1 + 2 \times \frac{1}{2}\right) = \left(-\frac{3}{2}, 2\right) \]
2. For the second end-point:
\[ (h + p, k - 2p) = \left(-2 + \frac{1}{2}, 1 - 2 \times \frac{1}{2}\right) = \left(-\frac{3}{2}, 0\right) \]

### Conclusion
The coordinates of the end-points of the rectum of the parabola are:
1. \((- \frac{3}{2}, 2)\)
2. \((- \frac{3}{2}, 0)\)