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)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

y2 = - 4(x - a2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/97884f463fb8db6ea2310f6c88242a1b-.png" /> Area = (1/2) (4a)(a2) = 2a3 Since 2a3 = 250 ⇒ 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

$y^{2} = - 4 (x - a^{2})$

Area $= \frac{1}{2} (4 a) (a^{2}) = 2 a^{3}$
Since $2 a^{3} = 250$
$\Rightarrow a = 5$

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

y2=−4(x−a2) Area =21​(4a)(a2)=2a3 Since 2a3=250 ⇒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

$y^{2} = - 4 (x - a^{2})$

Area $= \frac{1}{2} (4 a) (a^{2}) = 2 a^{3}$
Since $2 a^{3} = 250$
$\Rightarrow a = 5$