JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 16)
Let $$A\left( {{3 \over {\sqrt a }},\sqrt a } \right),\,a > 0$$, be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If $$D(3\cos \theta ,a\sin \theta )$$ is a point in the fourth quadrant such that the maximum area of $$\Delta$$ACD is 12 square units, then a is equal to ____________.
Answer
8
Explanation
Clearly $B$ is $\left(-\frac{3}{\sqrt{a}},+\sqrt{a}\right)$ and $C$ is $\left(-\frac{3}{\sqrt{a}},-\sqrt{a}\right)$
$$ \begin{aligned} &\text { Area of } \triangle A C D=\frac{1}{2}\left|\begin{array}{ccc} \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta| \\\\ &\Rightarrow \quad \Delta_{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^{2}=8 \end{aligned} $$
$$ \begin{aligned} &\text { Area of } \triangle A C D=\frac{1}{2}\left|\begin{array}{ccc} \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta| \\\\ &\Rightarrow \quad \Delta_{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^{2}=8 \end{aligned} $$
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