Key Concepts and Formulas

To effectively solve this problem, we will utilize the following fundamental concepts and formulas:

1. Standard Form of a Parabola: A parabola opening left or right has the standard form $(y-k)^2 = 4p(x-h)$, where $(h,k)$ is the vertex and $|p|$ is the distance from the vertex to the focus and directrix. If $4p$ is negative, the parabola opens to the left; if positive, it opens to the right.
2. Finding Intercepts: To find the points where a curve intersects the y-axis, we set the x-coordinate to zero ($x=0$) in its equation and solve for y. Similarly, for x-axis intercepts, we set $y=0$.
3. Area of a Triangle: While the general determinant formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ a much more efficient and less error-prone method, especially when one side of the triangle lies along a coordinate axis, is to use the formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ In this problem, two of the vertices lie on the y-axis, making this formula particularly convenient.

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Step 1: Determine the Vertex of the Parabola

The given equation of the parabola is $y^2 + 4(x - a^2) = 0$. Our first goal is to find the coordinates of its vertex, as this will be one of the triangle's vertices.

• Rearrange the equation into standard form: To identify the vertex easily, we need to express the equation in the standard form $(y-k)^2 = 4p(x-h)$. $y^2 = -4(x - a^2)$
• Compare with the standard form: By comparing $y^2 = -4(x - a^2)$ with $(y-k)^2 = 4p(x-h)$:
  • We can see that $k=0$ (since $y^2$ is equivalent to $(y-0)^2$).
  • We can see that $h=a^2$ (comparing $x-a^2$ with $x-h$).
  • The coefficient $4p = -4$, which implies $p=-1$. This tells us the parabola opens to the left.
• Identify the vertex: The vertex of the parabola is $(h,k)$. Therefore, the vertex of the given parabola is $V = (a^2, 0)$.

Explanation: Rewriting the equation in its standard form is crucial because it directly reveals the coordinates of the vertex $(h,k)$ and the orientation of the parabola. This makes the vertex's identification straightforward and accurate.

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Step 2: Find the Intersection Points with the y-axis

The other two vertices of the triangle are the points where the parabola intersects the y-axis.

• Condition for y-axis intersection: Any point on the y-axis has an x-coordinate of 0. So, to find these intersection points, we substitute $x=0$ into the parabola's equation. $y^2 = -4(0 - a^2)$
• Simplify and solve for y: $y^2 = -4(-a^2)$ $y^2 = 4a^2$ To find y, we take the square root of both sides: $y = \pm\sqrt{4a^2}$ $y = \pm 2a$
• Identify the intersection points: The two points where the parabola intersects the y-axis are $P_1 = (0, 2a)$ and $P_2 = (0, -2a)$.

Explanation: Setting $x=0$ is the standard procedure to find where any curve crosses the y-axis. This step provides the two remaining vertices of our triangle, which will lie symmetrically with respect to the x-axis due to the $y^2$ term in the parabola's equation.

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Step 3: Define the Vertices of the Triangle

Now we have successfully identified all three vertices that form the triangle:

• Vertex 1 (Parabola Vertex): $V = (a^2, 0)$
• Vertex 2 (y-intercept): $P_1 = (0, 2a)$
• Vertex 3 (y-intercept): $P_2 = (0, -2a)$

Let's denote the triangle as $\triangle VP_1P_2$.

Explanation: Clearly listing the vertices makes it easier to visualize the triangle and apply the area formula correctly in the next step.

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Step 4: Calculate the Area of the Triangle

We are given that the area of the triangle is 250 sq. units. We will calculate the area using the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.

• Identify the base: The points $P_1(0, 2a)$ and $P_2(0, -2a)$ both lie on the y-axis. This means the segment $P_1P_2$ forms a vertical line segment along the y-axis, making it an ideal choice for the base of our triangle. The length of the base $P_1P_2$ is the absolute difference of their y-coordinates: $\text{Base} = |2a - (-2a)| = |4a|$ Since the area is given as a positive value (250 sq. units), 'a' must be such that $4a$ is positive. Also, looking at the options, 'a' is positive. Therefore, we can simplify $\text{Base} = 4a$.

• Identify the height: The height of the triangle is the perpendicular distance from the third vertex, $V(a^2, 0)$, to the base $P_1P_2$, which lies on the y-axis. The perpendicular distance from any point $(x_0, y_0)$ to the y-axis is simply the absolute value of its x-coordinate, $|x_0|$. Thus, the height of the triangle is the x-coordinate of vertex $V$: $\text{Height} = |a^2| = a^2$ (Since $a^2$ is always non-negative).

• Calculate the Area: Now, substitute the base and height into the area formula: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area} = \frac{1}{2} \times (4a) \times (a^2)$ $\text{Area} = 2a^3$

Explanation: Choosing the base along the y-axis significantly simplifies the calculation of both the base length and the height. The base length is simply the difference in y-coordinates, and the height is the x-coordinate of the third vertex, which is the perpendicular distance to the y-axis. This method is much more efficient than using the general determinant formula for area.

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Step 5: Solve for 'a'

We are given that the area of the triangle is 250 sq. units. We can now equate our derived area expression with this given value.

• Set up the equation: $2a^3 = 250$
• Solve for $a^3$: Divide both sides by 2: $a^3 = \frac{250}{2}$ $a^3 = 125$
• Solve for $a$: Take the cube root of both sides: $a = \sqrt[3]{125}$ $a = 5$

Explanation: By equating the calculated area with the given area, we form a simple algebraic equation that allows us to determine the value of 'a'. Since we are looking for a real value of 'a' and $a^3 = 125$ is a positive number, there is a unique real solution for 'a'.

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Tips for Success and Common Pitfalls

• Tip 1: Always Convert to Standard Form: For parabolas, circles, ellipses, and hyperbolas, converting the given equation to its standard form is almost always the first and most crucial step. It immediately reveals key properties like the vertex, center, orientation, and focal length.
• Tip 2: Strategic Choice of Area Formula: When one side of a triangle lies along a coordinate axis (x-axis or y-axis), the formula $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ is vastly superior in terms of simplicity and reducing calculation errors compared to the determinant formula.
• Tip 3: Use Absolute Values for Distances: Remember that lengths (like base and height) are always positive. Use absolute values, e.g., $|x_2 - x_1|$ or $|y_2 - y_1|$, especially when variables are involved, to ensure positive results. In this specific problem, $a=5$ makes the terms positive naturally, but it's good practice.
• Common Mistake 1: Forgetting Both Roots: When solving an equation like $y^2 = K$, remember that $y = \pm\sqrt{K}$. Forgetting the negative root (e.g., just taking $y=2a$ instead of $y=\pm 2a$) would lead to incorrect vertices and an incorrect base length, resulting in an incorrect area.
• Common Mistake 2: Confusing Base and Height: Ensure you correctly identify which coordinate difference forms the base and which coordinate (or difference) forms the perpendicular height. For a base on the y-axis, the height is the absolute x-coordinate of the third vertex.

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Summary and Key Takeaway

This problem effectively tests your understanding of parabola properties and basic coordinate geometry. We systematically determined the vertices of the triangle by first transforming the parabola's equation into its standard form to find its vertex. Then, we found the y-intercepts by setting $x=0$. With the vertices established, we strategically used the $\frac{1}{2} \times \text{base} \times \text{height}$ formula for the triangle's area, choosing the segment on the y-axis as the base, which significantly simplified the calculations. Finally, we equated the derived area expression to the given area to solve for 'a'.

The key takeaway is that understanding the standard forms of conic sections and making strategic choices in geometric calculations (like selecting the most appropriate area formula) can greatly simplify complex problems.

The final answer is $\boxed{\text{5}}$.