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

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 sq. units, then a value of ' a ' is (A) 5 overline5 (B) (10)2 / 3 (C) 5(21 / 3) (D) 5

Tardigrade Admin · Accepted Answer

Vertex is ( a 2, 0) y2=-(x-a2) and x=0 ⇒(0, ± 2 a) Area of triangle is .=(1/2) .4 a .) a 2=250 ⇒ a 3=125 or a =5

Q. 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

$5 \overline{5}$

$(10)^{2/3}$

$5 (2^{1/3})$

$5$

Vertex is $(a^{2}, 0)$
$y^{2} = - (x - a^{2})$ and $x = 0 \Rightarrow (0, \pm 2 a)$
Area of triangle is $= \frac{1}{2} .4 a .) a^{2} = 250$
$\Rightarrow a^{3} = 125$ or $a = 5$

Q. 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

55

(10)2/3

5(21/3)

5

Vertex is (a2,0) y2=−(x−a2) and x=0⇒(0,±2a) Area of triangle is =21​.4a.)a2=250 ⇒a3=125 or a=5

Q. 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

$5 \overline{5}$

$(10)^{2/3}$

$5 (2^{1/3})$

$5$

Vertex is $(a^{2}, 0)$
$y^{2} = - (x - a^{2})$ and $x = 0 \Rightarrow (0, \pm 2 a)$
Area of triangle is $= \frac{1}{2} .4 a .) a^{2} = 250$
$\Rightarrow a^{3} = 125$ or $a = 5$