JEE PYQ: Parabola Question 32
Question 32 - 2019 (12 Jan Shift 2)
Let $P(4, -4)$ and $Q(9, 6)$ be two points on the parabola, $y^2 = 4x$ and let $X$ be any point on the arc POQ of this parabola, where $O$ is the vertex of this parabola, such that the area of $\Delta PXQ$ is maximum. Then this minimum area (in sq. units) is:
(1) $\frac{75}{2}$
(2) $\frac{125}{4}$
(3) $\frac{625}{4}$
(4) $\frac{125}{2}$
Show Answer
Answer: (2)
Solution
Let $X = (t^2, 2t)$. Area $= \frac{1}{2}|t^2(2t)(-4-6) + …|$. Using determinant: Area $= |-5t^2 + 5t + 30| = 5|t^2 - t - 6|$. For maximum, $t = \frac{1}{2}$: Area $= 5 \times \frac{25}{4} = \frac{125}{4}$.
BETA
