First slide

Cartesian plane

Question

$If A (\frac{3}{\sqrt{2}}, \sqrt{2}), B (\frac{- 3}{\sqrt{2}}, \sqrt{2}), C (\frac{- 3}{\sqrt{2}}, - \sqrt{2}) and D (3 \cos θ, 2 \sin θ) are four points. If$ $the area of quadrilateral ABCD is maximum (where θ \in (\frac{3 π}{2}, 2 π)), then$

Difficult

A

$maximum area is 10 sq. Units$
B

$θ = \frac{7 π}{4}$
C

$θ = 2 π - \sin^{- 1} (\frac{3}{\sqrt{85}})$
D

maximum area is 12 sq.units

Solution

Area of quadrilateral ABCD is maximum when area of ACD is maximum
Distance of D from AC $(2 x - 3 y = 0)$ is maximum $i.e., \cos θ - \sin θ is maximum$
$= \sqrt{2} \cos (θ + \frac{π}{4}) is maximum \Rightarrow θ = \frac{7 π}{4}$
$and area = \frac{6}{\sqrt{2}} \cdot 2 \sqrt{2} = 12 sq.units (since ABCD is a rectangle)$

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