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Byju's Answer

Standard XIII

Mathematics

Perpendicular Distance of a Point from a Line

If the area o...

Question

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 :

5 (2^{1 / 3})

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(10)^{2 / 3}

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5

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5 \sqrt{5}

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Solution

The correct option is C $5$
Consider the parabola

$Area (Δ A B C) = Area (Δ A O B) + Area (Δ A O C)$
$= \frac{1}{2} \times a^{2} \times 2 a + \frac{1}{2} \times a^{2} \times 2 a$
$= 2 a^{3}$
$∴ 2 a^{3} = 250$
$\Rightarrow a^{3} = 125$
$\Rightarrow a = 5$

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Similar questions

Q. The area (in sq. units) of an equilateral triangle inscribed in the parabola

y^{2} = 8 x

, with one of its vertices on the vertex of this parabola, is:

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y^{2} = 4 x

. If

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y^{2} = 4 x

and let

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of this parabola, where

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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 sq. units, then a value of 'a' is :

The correct option is C 5Consider the parabola Area(ΔABC)=Area(ΔAOB)+Area(ΔAOC) =12×a2×2a+12×a2×2a =2a3 ∴2a3=250 ⇒a3=125 ⇒a=5

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 :

The correct option is C $5$
Consider the parabola

$Area (Δ A B C) = Area (Δ A O B) + Area (Δ A O C)$
$= \frac{1}{2} \times a^{2} \times 2 a + \frac{1}{2} \times a^{2} \times 2 a$
$= 2 a^{3}$
$∴ 2 a^{3} = 250$
$\Rightarrow a^{3} = 125$
$\Rightarrow a = 5$