An equilateral triangle is inscribed in the parabola `y^2 = 4ax` whose vertex is at the vertex of the parabola. The length of its side is

Options

A. `2asqrt3`
B. `4asqrt3`
C. `6asqrt3`
D. `8asqrt3`

Answer

To find the length of the side of the equilateral triangle inscribed in the parabola \(y^2 = 4ax\) with its vertex at the vertex of the parabola, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
The parabola \(y^2 = 4ax\) opens to the right. The vertex of the parabola is at the origin (0, 0). We inscribe an equilateral triangle \(AOB\) with vertex \(O\) at the origin.

2. **Set Up Coordinates**:
Let the length of each side of the triangle be \(l\). The coordinates of points \(A\) and \(B\) can be expressed in terms of \(l\):
- Point \(A\) will be at \((l \cos 30^\circ, l \sin 30^\circ)\)
- Point \(B\) will be at \((l \cos 30^\circ, -l \sin 30^\circ)\)

3. **Calculate Coordinates**:
Using the values of \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):
- Coordinates of \(A\): \(\left(l \cdot \frac{\sqrt{3}}{2}, l \cdot \frac{1}{2}\right)\)
- Coordinates of \(B\): \(\left(l \cdot \frac{\sqrt{3}}{2}, -l \cdot \frac{1}{2}\right)\)

4. **Substitute into the Parabola Equation**:
Since point \(A\) lies on the parabola, we substitute its coordinates into the parabola equation \(y^2 = 4ax\):
\[ \left(l \cdot \frac{1}{2}\right)^2 = 4a \left(l \cdot \frac{\sqrt{3}}{2}\right) \]
Simplifying this gives:
\[ \frac{l^2}{4} = 4a \cdot \frac{l \sqrt{3}}{2} \]
\[ \frac{l^2}{4} = 2a l \sqrt{3} \]

5. **Rearranging the Equation**:
Multiply both sides by 4 to eliminate the fraction:
\[ l^2 = 8a l \sqrt{3} \]
Rearranging gives:
\[ l^2 - 8a l \sqrt{3} = 0 \]

6. **Factoring the Quadratic**:
Factor out \(l\):
\[ l(l - 8a \sqrt{3}) = 0 \]
This gives us two solutions: \(l = 0\) or \(l = 8a \sqrt{3}\). Since \(l = 0\) does not make sense in the context of a triangle, we have:
\[ l = 8a \sqrt{3} \]

7. **Conclusion**:
The length of each side of the inscribed equilateral triangle is \(8a \sqrt{3}\).