The extremities of latus rectum of a parabola are (1, 1) and `(1,-1)` . Then the equation of the parabola can be `y^2=2x-1` (b) `y^2=1-2x` `y^2=2x-3` (d) `y^2=2x-4`

Options

A. `y^(2)+2x-1`
B. `y^(2)=1-2x`
C. `y^(2)=3-2x`
D. `y^(2)=2x-4`

Explanation

To solve the problem, we need to find the equation of the parabola given the extremities of its latus rectum at the points (1, 1) and (1, -1).

### Step-by-step Solution:

1. **Identify the Extremities of the Latus Rectum:**
The extremities of the latus rectum are given as (1, 1) and (1, -1).

2. **Find the Midpoint:**
The midpoint of the latus rectum is the focus of the parabola. The midpoint can be calculated as:
\[ \text{Midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{1 + 1}{2}, \frac{1 + (-1)}{2}\right) = (1, 0) \]
Thus, the focus of the parabola is at (1, 0).

3. **Determine the Length of the Latus Rectum:**
The distance between the extremities of the latus rectum is:
\[ \text{Length} = |y_1 - y_2| = |1 - (-1)| = 2 \]
The length of the latus rectum is given by \(4a\), where \(a\) is the distance from the vertex to the focus. Therefore, we have:
\[ 4a = 2 \implies a = \frac{1}{2} \]

4. **Find the Vertex:**
Since the focus is at (1, 0) and the parabola opens horizontally (as the y-coordinates of the latus rectum are different but the x-coordinates are the same), the vertex will be located \(a\) units to the left of the focus. Thus, the vertex is:
\[ \text{Vertex} = (1 - a, 0) = (1 - \frac{1}{2}, 0) = \left(\frac{1}{2}, 0\right) \]

5. **Write the Equation of the Parabola:**
The standard form of the equation of a parabola that opens to the right is:
\[ (y - k)^2 = 4a(x - h) \]
Here, \((h, k)\) is the vertex \(\left(\frac{1}{2}, 0\right)\) and \(4a = 2\). Substituting these values into the equation gives:
\[ (y - 0)^2 = 2\left(x - \frac{1}{2}\right) \]
Simplifying this, we get:
\[ y^2 = 2x - 1 \]

6. **Conclusion:**
The equation of the parabola is:
\[ y^2 = 2x - 1 \]

### Final Answer:
The correct option is (a) \(y^2 = 2x - 1\).