A square has one vertex at the vertex of the parabola `y^2=4a x` and the diagonal through the vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinates of the vertices of the square are (a)`(4a ,4a)` (b) `(4a ,-4a)` (c)`(0,0)` (d) `(8a ,0)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Parabola
The equation of the parabola is given as \( y^2 = 4ax \). The vertex of this parabola is at the origin \( (0, 0) \).

### Step 2: Identify the Square's Properties
The square has one vertex at the vertex of the parabola, which is \( (0, 0) \). The diagonal through this vertex lies along the x-axis. This means that the square is oriented such that its sides are parallel to the axes.

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