JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Parabola

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) $55$
(B) $5 (2^{1/3})$
(C) $(10)^{2/3}$
(D)5✓ Correct

Answer: (D)

Solution

The parabola

y^{2} = - 4 (x - a^{2})

has its vertex at

(a^{2}, 0)

and meets the

y

-axis where

y^{2} = 4 a^{2}

, i.e. at

(0, 2 a)

and

(0, - 2 a)

. The triangle with these three corners has base

4 a

along the

y

-axis and height

a^{2}

, so its area is

\frac{1}{2} \times 4 a \times a^{2} = 2 a^{3}

. Setting

2 a^{3} = 250

gives

a^{3} = 125

and

a = 5

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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