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Step-by-Step Solution
Step 1: Write down the given equation of the parabola
The parabola is given by
$y^2 + 4(x - a^2) = 0$.
We can rewrite it as
$y^2 = -4\bigl(x - a^2\bigr)$.
Step 2: Identify the vertex of the parabola
From the rewritten form $y^2 = -4\bigl(x - a^2\bigr)$, we see that the vertex occurs at
$x = a^2$ (where $y=0$).
Thus, the vertex is
$(a^2,\, 0)$.
Step 3: Find the intersection of the parabola with the y-axis
The y-axis is where $x = 0$. Plug $x = 0$ into the parabola’s equation:
$y^2 = -4\bigl(0 - a^2\bigr) = 4a^2 \quad \Longrightarrow \quad y = \pm\,2a.$
Hence, the points of intersection with the y-axis are
$(0,\, 2a)$ and $(0,\, -2a)$.
Step 4: Determine the vertices of the triangle and its base and height
We have a triangle whose vertices are:
$\bigl(a^2,\,0\bigr)$ (the vertex of the parabola)
$\bigl(0,\,2a\bigr)$
$\bigl(0,\,-2a\bigr)$
The distance between the points on the y-axis, $(0,2a)$ and $(0,-2a)$, is the base:
$ \text{Base} = 2a - (-2a) = 4a.$
The height of this triangle is the horizontal distance from $(a^2,0)$ to the y-axis ($x=0$). Thus,
$ \text{Height} = a^2.$
Step 5: Use the area formula of a triangle
The area of a triangle is given by:
$$
\text{Area} = \frac{1}{2}\,(\text{base})\,(\text{height}).
$$
Substituting $\text{base} = 4a$ and $\text{height} = a^2$, we get:
$$
\text{Area} = \frac{1}{2} \times 4a \times a^2 = 2\,a^3.
$$
Step 6: Equate to the given area and solve for a
We are told that the area is $250$ square units. Therefore,
$$
2\,a^3 = 250
\quad \Longrightarrow \quad
a^3 = 125
\quad \Longrightarrow \quad
a = 5.
$$
Step 7: Conclude the value of a
The value of $a$ that satisfies the given condition is
$\boxed{5}.$
