The maximum area of a triangle whose one vertex is at ( 0 , 0 ) and the other two vertices lie on the curve…

Let, area of triangle $ABC$ be denoted by $P$.

$\Rightarrow P=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ x& y& 1\\ -x& y& 1\end{array}\right|$

$\Rightarrow P=\frac{1}{2}\left|0-0+2xy\right|$

$\Rightarrow P=\left|xy\right|$

$\Rightarrow P=\left|x\left(-2x^2+54\right)\right|$

$\Rightarrow P=\left|-2x^3+54x\right|$

$\Rightarrow \frac{dP}{dx}=\left|-6x^2+54\right|$

For critical points, $\frac{dP}{dx}=0$

$\Rightarrow \left|-6x^2+54\right|=0$

$\Rightarrow x=\pm 3$

So, the maximum area will be,

$A=3(-2\times 9+54)$

$\Rightarrow A=108$