If the area of the triangle whose one vertex is at the vertex of the parabola,

Question:

If the area of the triangle whose one vertex is at the vertex of the parabola, $y^{2}+4\left(x-a^{2}\right)=0$ 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 :-

  1. $5 \sqrt{5}$

  2. $(10)^{2 / 3}$

  3. $5\left(2^{1 / 3}\right)$

  4. 5


Correct Option: , 4

Solution:

Vertex is $\left(a^{2}, 0\right)$

$\mathrm{y}^{2}=-\left(\mathrm{x}-\mathrm{a}^{2}\right)$ and $\mathrm{x}=0 \Rightarrow(0, \pm 2 \mathrm{a})$

Area of triangle is $=\frac{1}{2}, 4 a \cdot\left(a^{2}\right)=250$

$\Rightarrow a^{3}=125$ or $a=5$

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