1. Write the parabola in standard form

Given: $y^2 + 4(x-a^2)=0$

So, $y^2=-4(x-a^2)$

This is a parabola of the form: $y^2=-4(x-h)$ with vertex at $(h,0)=(a^2,0)$

Hence, the vertex is: $V=(a^2,0)$

2. Find the points where the parabola cuts the y-axis

On the y-axis, $x=0$.

Substitute $x=0$ into the parabola: $y^2+4(0-a^2)=0$ $y^2-4a^2=0$ $y^2=4a^2$ $y=\pm 2a$

So the points of intersection are: $P=(0,2a), \qquad Q=(0,-2a)$

3. Form the triangle and compute its area

The triangle has vertices: $V=(a^2,0),\quad P=(0,2a),\quad Q=(0,-2a)$

Segment $PQ$ lies on the y-axis, so its length is: $PQ=2a-(-2a)=4a$

The perpendicular distance of the vertex $V=(a^2,0)$ from the y-axis is: $a^2$

Therefore, area of the triangle is: $\text{Area}=\frac12\times PQ\times \text{distance from }V\text{ to y-axis}$ $=\frac12\times 4a\times a^2$ $=2a^3$

Given area is $250$ sq. units, so: $2a^3=250$ $a^3=125$ $a=5$

4. Check the options

• A: $5\sqrt5$ ✗
• B: $(10)^{2/3}$ ✗
• C: $5\,2^{1/3}$ ✗
• D: $5$ ✓

Thus, the correct option is: $\boxed{D}$