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JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Parabola

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Parabola

If the area of the triangle whose one vertex is at the vertex of...

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 : 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

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

Options

A $5 \sqrt{5}$
B $5\left(2^{1 / 3}\right)$
C $(10)^{33}$
D5

Correct answer

D. 5

Step-by-step solution

y^{t}=-4

\left(x-a^{2}\right)

<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/0pFWq6xYy6euRwfWHifO2c-ZqETw7H3Bt9DZMpw2PCI.original.fullsize.png"><br>

\mathrm{Area}=\frac{1}{2}(4 a)\left(a^{2}\right)=2 a^{3}

Since

2 a^{3}=250 \Rightarrow a=5

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