If a0, b0 then the maximum area of the parallelogram whose three vertices are O(0,0) ⋅ A(a cos θ, b sin θ) and B(a cos θ,-b sin θ) is

Question

If a>0, b>0 then the maximum area of the parallelogram whose three vertices are O(0,0) ⋅ A(a cos θ, b sin θ) and B(a cos θ,-b sin θ) is (A) ab when θ=π / 4 (B) 3 a b when θ=π / 4 (C) ab when θ=π / 2 (D) 2 a b

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b52d8e491e1d95706a7070ee9f975345-.png" /> Area O A B C=2 × area triangle O A B =2 ×((1/2) × O M × A B) =2 × (1/2)|a cos θ × 2 b sin θ| =2 × a b| sin θ cos θ| =a b| sin 2 θ| ∴ Maximum is ab when θ=π / 4

Q. If $a > 0, b > 0$ then the maximum area of the parallelogram whose three vertices are $O (0, 0) \cdot A (a cos θ, b sin θ)$ and $B (a cos θ, - b sin θ)$ is

$ab$ when $θ = π /4$

$3 ab$ when $θ = π /4$

$ab$ when $θ = π /2$

$2 ab$

Area $O A BC = 2 \times$ area $△ O A B$
$= 2 \times (\frac{1}{2} \times OM \times A B)$
$= 2 \times \frac{1}{2} ∣ a cos θ \times 2 b sin θ ∣$
$= 2 \times ab ∣ sin θ cos θ ∣$
$= ab ∣ sin 2 θ ∣$
$∴$ Maximum is ab when $θ = π /4$

Q. If a>0,b>0 then the maximum area of the parallelogram whose three vertices are O(0,0)⋅A(acosθ,bsinθ) and B(acosθ,−bsinθ) is

ab when θ=π/4

3ab when θ=π/4

ab when θ=π/2

2ab