All the three vertices of an equilateral triangle lie on the parabola `y = x^(2)`, and one of its sides has a slope of 2. Then the sum of the x-coordinates of the three vertices is

Options

A. `(5)/(9)`
B. `(9)/(13)`
C. `(6)/(11)`
D. None of these

Explanation

To solve the problem, we need to find the sum of the x-coordinates of the three vertices of an equilateral triangle that lies on the parabola \( y = x^2 \) and has one side with a slope of 2.

### Step-by-Step Solution:

1. **Understanding the Parabola and Triangle**:
The parabola is given by \( y = x^2 \). The vertices of the equilateral triangle will be of the form \( (t_1, t_1^2) \), \( (t_2, t_2^2) \), and \( (t_3, t_3^2) \).

2. **Finding the Slope of the Side**:
One side of the triangle has a slope of 2. The slope between two points \( (t_1, t_1^2) \) and \( (t_2, t_2^2) \) can be calculated as:
\[ \text{slope} = \frac{t_2^2 - t_1^2}{t_2 - t_1} \]
Using the difference of squares, this simplifies to:
\[ \text{slope} = \frac{(t_2 - t_1)(t_2 + t_1)}{t_2 - t_1} = t_2 + t_1 \]
Setting this equal to 2 gives us:
\[ t_1 + t_2 = 2 \quad \text{(Equation 1)} \]

3. **Finding the Third Vertex**:
The third vertex \( (t_3, t_3^2) \) will also form slopes with the other two vertices. The angles in an equilateral triangle are all \( 60^\circ \). Therefore, we can use the slope conditions to find relationships between \( t_1, t_2, \) and \( t_3 \).

4. **Using the Slope Condition**:
The slope between \( (t_1, t_1^2) \) and \( (t_3, t_3^2) \) is:
\[ \text{slope} = \frac{t_3^2 - t_1^2}{t_3 - t_1} = t_3 + t_1 \]
Setting this equal to the tangent of \( 60^\circ \), which is \( \sqrt{3} \):
\[ t_3 + t_1 = \sqrt{3} \quad \text{(Equation 2)} \]

5. **Finding the Third Slope**:
Similarly, the slope between \( (t_2, t_2^2) \) and \( (t_3, t_3^2) \) is:
\[ t_3 + t_2 = \sqrt{3} \quad \text{(Equation 3)} \]

6. **Solving the System of Equations**:
We now have three equations:
\[ t_1 + t_2 = 2 \quad (1) \]
\[ t_3 + t_1 = \sqrt{3} \quad (2) \]
\[ t_3 + t_2 = \sqrt{3} \quad (3) \]

From (2), we can express \( t_3 \):
\[ t_3 = \sqrt{3} - t_1 \]
From (3):
\[ t_3 = \sqrt{3} - t_2 \]

Setting the two expressions for \( t_3 \) equal gives:
\[ \sqrt{3} - t_1 = \sqrt{3} - t_2 \implies t_1 = t_2 \]

7. **Substituting Back**:
Substitute \( t_1 = t_2 \) into Equation (1):
\[ 2t_1 = 2 \implies t_1 = 1 \quad \text{and hence} \quad t_2 = 1 \]

Now substituting \( t_1 \) into Equation (2) to find \( t_3 \):
\[ t_3 + 1 = \sqrt{3} \implies t_3 = \sqrt{3} - 1 \]

8. **Calculating the Sum of x-coordinates**:
The sum of the x-coordinates of the vertices is:
\[ t_1 + t_2 + t_3 = 1 + 1 + (\sqrt{3} - 1) = 1 + \sqrt{3} \]

### Final Answer:
The sum of the x-coordinates of the three vertices is \( 1 + \sqrt{3} \).