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 √5 (B) (10)2/3 (C) 5 (21/3) (D) 5

Tardigrade Admin · Accepted Answer

Vertex is (a2,0) y2 = -(x - a2) and x = 0 ⇒ (0, ± 2a) Area of triangle is = (1/2) . 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 $250 s q$ . units, then a value of $^{'} a^{'}$ is :

$55$

$(10)^{2/3}$

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

$5$

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 250sq. 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 $250 s q$ . units, then a value of $^{'} a^{'}$ is :

$55$

$(10)^{2/3}$

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

$5$

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 $a = 5$