The extremities of latus rectum of the parabola `(y - 1)^2 = 2(x+2)` are

Options

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

Answer

To find the extremities of the latus rectum of the parabola given by the equation \((y - 1)^2 = 2(x + 2)\), we will follow these steps:

### Step 1: Identify the standard form of the parabola
The given equation can be rewritten in a more recognizable form. The standard form of a parabola that opens to the right is given by:

\[ (y - k)^2 = 4a(x - h) \]

where \((h, k)\) is the vertex of the parabola.

### Step 2: Rewrite the equation
We can rewrite the given equation:

\[ (y - 1)^2 = 2(x + 2) \]

This can be rearranged to match the standard form:

\[ (y - 1)^2 = 2(x + 2) \implies (y - 1)^2 = 2x + 4 \]

### Step 3: Identify the vertex and parameter \(a\)
From the equation \((y - 1)^2 = 2(x + 2)\), we can identify:
- The vertex \((h, k)\) is \((-2, 1)\).
- Here, \(4a = 2\), which gives us \(a = \frac{2}{4} = \frac{1}{2}\).

### Step 4: Find the extremities of the latus rectum
The extremities of the latus rectum of a parabola are given by the points:

\[ (h + a, k \pm 2a) \]

Substituting the values we found:
- \(h = -2\)
- \(k = 1\)
- \(a = \frac{1}{2}\)

Calculating the coordinates:

1. **X-coordinate**:
\[ h + a = -2 + \frac{1}{2} = -2 + 0.5 = -1.5 = -\frac{3}{2} \]

2. **Y-coordinates**:
\[ k \pm 2a = 1 \pm 2 \times \frac{1}{2} = 1 \pm 1 \]
This gives us two values:
- \(1 + 1 = 2\)
- \(1 - 1 = 0\)

### Step 5: Write the final coordinates
Thus, the extremities of the latus rectum are:

\[ \left(-\frac{3}{2}, 2\right) \quad \text{and} \quad \left(-\frac{3}{2}, 0\right) \]

### Final Answer:
The extremities of the latus rectum of the parabola \((y - 1)^2 = 2(x + 2)\) are \(\left(-\frac{3}{2}, 2\right)\) and \(\left(-\frac{3}{2}, 0\right)\).

---