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QJEEM22Mad2c2: JEE Main 2022 Mathematics - Class 11 JEEM22Mad2c2 Solution: Let , be a fixed point in the xy-plane. The image of A in…

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(3cos⁡θ,asin⁡θ)D(3\cos \theta ,a\sin \theta ) is a point in the fourth quadrant such that the maximum area of Δ\Delta ACD is 12 square units, then a is equal to ____________.

Step-by-Step Verified Solution

Clearly BB is (−3a,+a)\left(-\frac{3}{\sqrt{a}},+\sqrt{a}\right) and CC is (−3a,−a)\left(-\frac{3}{\sqrt{a}},-\sqrt{a}\right)

amp; Area of △ACD=12∣3aamp;aamp;1−3aamp;−aamp;13cos⁡θamp;asin⁡θamp;1∣amp;⇒Δ=∣0amp;0amp;1−3aamp;−aamp;13cos⁡θamp;asin⁡θamp;1∣amp;⇒Δ=∣3asin⁡θ+3acos⁡θ∣=3a∣sin⁡θ+cos⁡θ∣amp;⇒Δmax⁡=3a⋅2=12⇒a=(22)2=8\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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