JEE PYQ: Parabola Question 23
Question 23 - 2019 (09 Jan Shift 2)
Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola, $y^2 = 4x$. Let $C$ be chosen on the arc AOB of the parabola, where $O$ is the origin, such that the area of $\Delta ACB$ is maximum. Then, the area (in sq. units) of $\Delta ACB$, is:
(1) $31\frac{1}{4}$
(2) $30\frac{1}{2}$
(3) $32$
(4) $31\frac{3}{4}$
Show Answer
Answer: (1)
Solution
Let $C = (t^2, 2t)$. Area $= \frac{1}{2}|t^2(6+4) - 2t(9-4) + 1(-36-24)| = |5t^2 - 5t - 30| = 5|t^2 - t - 6|$. For $t \in (0, 3)$: $= 5\left|\left(t - \frac{1}{2}\right)^2 - \frac{25}{4}\right|$. Maximum at $t = \frac{1}{2}$: area $= 5 \times \frac{25}{4} = \frac{125}{4} = 31\frac{1}{4}$.
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