JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 5)
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 :
108
122
88
92
Explanation
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Area of $$\Delta$$
$$\begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1 \end{array}\right| \\ & \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\ & \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\ & \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \\ & \text { Area }=3(-2 \times 9+54)=108 \end{aligned}$$
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