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 other two 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. $5 (2^{1/3})$
• D. $5$

Answer 1

Answer: (D)

Vertex is $(a^2,0)$
$y^2 = -(x - a^2)$ and $x = 0 \; \Rightarrow \; (0, \pm 2a)$
Area of triangle is $= \frac{1}{2} . 4a.(a^2) = 250$
$\Rightarrow \; a^3 = 125$ or $a = 5$