Let A (3 a , a ) a >0, be a fixed point in the xy-plane. The imag… - Vidyadip

MCQ

Let $A \left(\frac{3}{\sqrt{ a }}, \sqrt{ a }\right) a >0$, 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 $\triangle ACD$ is $12$ square units, then $a$ is equal to

A
$12$
✓
$8$
C
$6$
D
$3$

✓

Answer

Correct option: B.

$8$

b
$A =\left(\frac{3}{\sqrt{ a }}, \sqrt{ a }\right)$

$B =\left(\frac{-3}{\sqrt{ a }}, \sqrt{ a }\right)$

$C =\left(-\frac{3}{\sqrt{ a }},-\sqrt{ a }\right)$

Area of $ACD$

$\frac{1}{2}\left|\begin{array}{cc}\frac{3}{\sqrt{a}} & \sqrt{a} \\ -\frac{3}{\sqrt{a}} & -\sqrt{a} \\ 3 \cos \theta & a \sin \theta \\ \frac{3}{\sqrt{a}} & \sqrt{a}\end{array}\right|$

$\frac{1}{2} 6 \sqrt{ a }(\cos \theta-\sin \theta)$

$3 \sqrt{ a }(\cos \theta-\sin \theta)$

max values of function is $3 \sqrt{ a } \sqrt{2}$

$3 \sqrt{a} \sqrt{2}=12$

$2 a=16$

$a=8$

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