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 $y>0$ is :
(1) 88
(2) 122
(3) 92
(4) 108
Show Answer
Answer (4)
Solution
Area of $\Delta$
$=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1\end{array}\right|$
$\Rightarrow\left|\frac{1}{2}(xy+xy)\right|=|xy|$
$\operatorname{Area}(\Delta)=|xy|=\left|x\left(-2 x^{2}+54\right)\right|$
$\frac{d(\Delta)}{dx}=\left|\left(-6 x^{2}+54\right)\right| \Rightarrow \frac{d \Delta}{dx}=0$ at $x=3$
Area $=3(-2 \times 9+54)=108$
