If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2)= 0 and the other two vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is:

To find the value of 'a' in the given problem, we need to follow these steps:

Step 1: Write the equation of the parabola in standard form. The given equation, y^2 + 4(x - a^2) = 0, is already in standard form.

Step 2: Find the points of intersection of the parabola and the y-axis. To find the points of intersection, we set x = 0 in the equation of the parabola. So, y^2 + 4(0 - a^2) = 0 Simplifying, we get y^2 - 4a^2 = 0 Factoring, we have (y - 2a)(y + 2a) = 0 So, y = 2a or y = -2a

Step 3: Find the coordinates of the vertices of the triangle. The vertex of the parabola is at (a^2, 0). The other two vertices are the points of intersection with the y-axis, which are (0, 2a) and (0, -2a).

Step 4: Calculate the area of the triangle. The area of a triangle can be calculated using the formula: Area = (1/2) * base * height. In this case, the base of the triangle is the distance between the two points of intersection on the y-axis, which is 2a - (-2a) = 4a. The height of the triangle is the distance between the vertex of the parabola and the y-axis, which is a^2. So, the area of the triangle is (1/2) * 4a * a^2 = 2a^3.

Step 5: Set up the equation and solve for 'a'. Given that the area of the triangle is 250 sq. units, we have: 2a^3 = 250 Dividing both sides by 2, we get: a^3 = 125 Taking the cube root of both sides, we find: a = 5

Therefore, the value of 'a' is 5.