A point P on the parabola `y^(2)=4x`, the foot of the perpendicular from it upon the directrix and the focus are the vertices of an equilateral triangle. If the area of the equilateral triangle is `beta` sq. units, then the value of `beta^(2)` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the parabola and its properties
The given parabola is \( y^2 = 4x \). From this, we can identify that \( a = 1 \).

### Step 2: Determine the directrix and focus
For the parabola \( y^2 = 4ax \):
- The focus \( F \) is at \( (a, 0) = (1, 0) \).
- The directrix is given by the equation \( x = -a = -1 \).

### Step 3: Define a point \( P \) on the parabola
Let \( P \) be a point on the parabola. The coordinates of point \( P \) can be expressed in terms of a parameter \( t \):
\[ P(t) = (t^2, 2t) \]

### Step 4: Find the foot of the perpendicular from \( P \) to the directrix
The foot of the perpendicular from point \( P \) to the directrix \( x = -1 \) will have coordinates \( M(-1, 2t) \).

### Step 5: Set up the triangle vertices
The vertices of the equilateral triangle are:
- \( P(t) = (t^2, 2t) \)
- \( F = (1, 0) \)
- \( M = (-1, 2t) \)

### Step 6: Calculate the lengths of the sides of the triangle
We will calculate the lengths of the sides \( PF \), \( FM \), and \( MP \):
1. **Length \( PF \)**:
\[ PF = \sqrt{(t^2 - 1)^2 + (2t - 0)^2} = \sqrt{(t^2 - 1)^2 + 4t^2} \]
Simplifying:
\[ PF = \sqrt{t^4 - 2t^2 + 1 + 4t^2} = \sqrt{t^4 + 2t^2 + 1} = \sqrt{(t^2 + 1)^2} = t^2 + 1 \]

2. **Length \( FM \)**:
\[ FM = \sqrt{(-1 - 1)^2 + (2t - 0)^2} = \sqrt{(-2)^2 + (2t)^2} = \sqrt{4 + 4t^2} = 2\sqrt{1 + t^2} \]

3. **Length \( MP \)**:
\[ MP = \sqrt{(t^2 + 1)^2 + (2t - 2t)^2} = t^2 + 1 \]

### Step 7: Set the lengths equal for an equilateral triangle
Since \( PFM \) is an equilateral triangle, we have:
\[ PF = FM = MP \]
From our calculations:
\[ t^2 + 1 = 2\sqrt{1 + t^2} \]

### Step 8: Solve for \( t \)
Squaring both sides:
\[ (t^2 + 1)^2 = 4(1 + t^2) \]
Expanding:
\[ t^4 + 2t^2 + 1 = 4 + 4t^2 \]
Rearranging gives:
\[ t^4 - 2t^2 - 3 = 0 \]
Let \( u = t^2 \):
\[ u^2 - 2u - 3 = 0 \]
Factoring:
\[ (u - 3)(u + 1) = 0 \]
Thus, \( u = 3 \) or \( u = -1 \) (discard \( u = -1 \) since \( t^2 \) cannot be negative).

So, \( t^2 = 3 \).

### Step 9: Calculate the area of the triangle
The side length \( s \) of the equilateral triangle is:
\[ s = PF = t^2 + 1 = 3 + 1 = 4 \]
The area \( A \) of an equilateral triangle is given by:
\[ A = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \cdot 4^2 = \frac{\sqrt{3}}{4} \cdot 16 = 4\sqrt{3} \]

### Step 10: Find \( \beta^2 \)
Thus, \( \beta = 4\sqrt{3} \) and:
\[ \beta^2 = (4\sqrt{3})^2 = 16 \cdot 3 = 48 \]

### Final Answer
The value of \( \beta^2 \) is \( \boxed{48} \).