Question

If the area of the triangle whose one vertex is at the vertex of the parabola, \(y^2 + 4(x - a^2) = 0\) and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :

• A. \(5\sqrt{5}\)
• B. \((10)^{2/3}\)
• C. \(5\left(2^{1/3}\right)\)
• D. 5

Hint:

The vertex is (a^2, 0). Put x = 0 to get the two y-axis points (0, 2a) and (0, -2a), then use base times height.

Answer 1

The Correct Option is D

The parabola is \(y^2 = -4(x - a^2)\), with vertex \((a^2, 0)\).
Intersection with the y-axis (x = 0): \(y^2 = 4a^2\), so \(y = \pm 2a\). The points are \((0, 2a)\) and \((0, -2a)\).
Triangle: base on the y-axis of length \(4|a|\), height = distance of the vertex from the y-axis = \(a^2\).
Area = \(\frac{1}{2}\cdot 4|a|\cdot a^2 = 2|a|^3 = 250\), so \(|a|^3 = 125\) and \(|a| = 5\).
A value of a is 5.
Answer: (4) 5