Q39 / 90 · JEE Main 2024 (30 Jan Shift 1) · Mathematics

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 :

Solution

Let, area of triangle $A B C$ be denoted by $P$ .

$\Rightarrow P = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ x & y & 1 \\ - x & y & 1 \end{matrix}|$

$\Rightarrow P = \frac{1}{2} |0 - 0 + 2 x y|$

$\Rightarrow P = |x y|$

$\Rightarrow P = |x (- 2 x^{2} + 54)|$

$\Rightarrow P = |- 2 x^{3} + 54 x|$

$\Rightarrow \frac{d P}{d x} = |- 6 x^{2} + 54|$

For critical points, $\frac{d P}{d x} = 0$

$\Rightarrow |- 6 x^{2} + 54| = 0$

$\Rightarrow x = \pm 3$

So, the maximum area will be,

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

$\Rightarrow A = 108$

Hence, the correct answer is option: 'D'.