Parabola Question 3
Question 3 - 2024 (30 Jan Shift 1)
The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y=-2 x^{2}+54$ at points $(x, y)$ and $(-x, y)$ where $\mathrm{y}>0$ is :
(1) 88
(2) 122
(3) 92
(4) 108
Show Answer
Answer (4)
Solution
Area of $\Delta$
$=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \ \mathrm{x} & \mathrm{y} & 1 \ -\mathrm{x} & \mathrm{y} & 1\end{array}\right|$
$\Rightarrow\left|\frac{1}{2}(\mathrm{xy}+\mathrm{xy})\right|=|\mathrm{xy}|$
$\operatorname{Area}(\Delta)=|\mathrm{xy}|=\left|\mathrm{x}\left(-2 \mathrm{x}^{2}+54\right)\right|$
$\frac{\mathrm{d}(\Delta)}{\mathrm{dx}}=\left|\left(-6 \mathrm{x}^{2}+54\right)\right| \Rightarrow \frac{\mathrm{d} \Delta}{\mathrm{dx}}=0$ at $\mathrm{x}=3$
Area $=3(-2 \times 9+54)=108$
