Let $\mathrm{A}\left(\frac{3}{\sqrt{\mathrm{a}}}, \sqrt{\mathrm{a}}\right) \mathrm{a}>0$, be a fixed point in the $x y$-plane. The image of $A$ in $y$-axis be $B$ and the image of B in x -axis be C . If $\mathrm{D}(3 \cos \theta, a \sin \theta)$ is a point in the fourth quadrant such that the maximum area of $\triangle \mathrm{ACD}$ is 12 square units, then a is equal to \_\_\_\_ .

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Straight Lines

Let

A (\frac{3}{a}, a) a > 0

, be a fixed point in the

x y

-plane. The image of

A

y

-axis be

B

and the
image of B in x -axis be C . If

D (3 cos θ, a sin θ)

is a point in the fourth quadrant such that the maximum area of

△ ACD

is 12 square units, then a is equal to ____ .

Answer: 8

Solution

Given

A = (\frac{3}{a}, a)

, its reflection in the y-axis is

B = (- \frac{3}{a}, a)

, and B's reflection in the x-axis is

C = (- \frac{3}{a}, - a)

.

Using the determinant formula for area:

Area = \frac{1}{2} \frac{3}{a} (a sin θ + a) - a (3 cos θ + \frac{3}{a}) - \frac{3}{a} (a sin θ - a)

Simplifying:

Area = 3 a ∣ cos θ - sin θ ∣

.

Maximum of

∣ cos θ - sin θ ∣

2

, so max area =

3 2 a

.

Given max area = 12:

3 2 a = 12

, so

2 a = 4

, giving

2 a = 16

, thus

a = 8

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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