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 s units, then a value of ‘a’ is:

Engineering

Mathematics

Area of a Triangle

Standard Parabola

Question

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 s units, then a value of ‘a’ is:

5(2^1/3)

$5 \sqrt{5}$

(10)^2/3

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Vertices of the $∆$ are (0, 2a), (0, – 2a) and (a², 0)

Area = $\frac{1}{2} | 4 a^{3} |$ = 250 ⇒ 5