JEE PYQ: Parabola Question 27
Question 27 - 2019 (11 Jan Shift 2)
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:
(1) $5\sqrt{5}$
(2) $5(2^{1/3})$
(3) $(10)^{2/3}$
(4) $5$
Show Answer
Answer: (4)
Solution
$y^2 = -4(x - a^2)$. Vertex $(a^2, 0)$. Intersection with $y$-axis: $y^2 = 4a^2 \Rightarrow y = \pm 2a$. Points: $(0, 2a)$ and $(0, -2a)$. Area $= \frac{1}{2}(4a)(a^2) = 2a^3 = 250 \Rightarrow a = 5$.
BETA
