JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 19)
If the area of the triangle whose one vertex is at the vertex of the parabola, y2 + 4(x – a2) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is :
$$5\sqrt 5 $$
$${\left( {10} \right)^{2/3}}$$
$$5\left( {{2^{1/3}}} \right)$$
5
Explanation
Vertex is (a2, 0)
y2 $$=$$ $$-$$(x $$-$$ a2) and x $$=$$ 0 $$ \Rightarrow $$ (0, $$ \pm $$ 2a)
Area of triangle is $$ = {1 \over 2}.$$4a.(a2) = 250
$$ \Rightarrow $$ a3 = 125 or a = 5
y2 $$=$$ $$-$$(x $$-$$ a2) and x $$=$$ 0 $$ \Rightarrow $$ (0, $$ \pm $$ 2a)
Area of triangle is $$ = {1 \over 2}.$$4a.(a2) = 250
$$ \Rightarrow $$ a3 = 125 or a = 5
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