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

Answer 1

The correct option is C $5$

Consider the parabola

$\text{Area}\left(\Delta ABC\right)=\text{Area}\left(\Delta AOB\right)+\text{Area}\left(\Delta AOC\right)$
$=\frac{1}{2}\times a^2\times 2a+\frac{1}{2}\times a^2\times 2a$
$=2a^3$
$\therefore 2a^3=250$
$\Rightarrow a^3=125$
$\Rightarrow a=5$