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 :

Parabola: If the area of the triangle whose one vertex is at the vertex of the… | JEE Main 2019

JEE Main 2019 (Online) 11th January Evening ShiftQuestion 1

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 :

Parabola

$5\sqrt5$$5\left(2^{1/3}\right)$$(10)^{33}$$5$✓ Correct answer

Solution

Concept. Find the vertex and the two points on the $y$-axis, then use base $\times$ height. Formulas used. (1) Area $=\frac12\times$ base $\times$ height.Step 1 - Vertex. $y^2=-4(x-a^2)$ has vertex $(a^2,0)$. Step 2 - On the $y$-axis. $x=0$: $y^2=4a^2\Rightarrow y=\pm2a$. Base $=4|a|$. Step 3 - Area (1). Height $=a^2$ (distance of the vertex from the $y$-axis): $$\frac12\cdot4|a|\cdot a^2=2|a|^3=250$$ Step 4 - Solve. $|a|^3=125\Rightarrow|a|=5$, so $a=5$ is a value. Answer: D

Parabola - JEE Mathematics question (JEE Main 2019 (Online) 11th January Evening Shift)

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