The area of the greatest rectangle that can be inscribed in the ellipse (x2/a2)+(y2/b2)=1 is

Question

The area of the greatest rectangle that can be inscribed in the ellipse (x2/a2)+(y2/b2)=1 is (A) ab (B) 2 ab (C) ( a / b ) (D) √ ab

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/94b60cafa95f50bc79306d15fbb0cca3-.png" /> Let P Q R S be a rectangle, where P is (a cos θ, b sin θ) ∴ text Area of rectangle =4(a cos θ),(b sin θ) =2 a b sin 2 θ This is maximum when sin 2 θ=1 Hence, maximum area =2 ab (1)=2 ab

Q. The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is

ab

$2 ab$

$\frac{a}{b}$

$ab$

Let $PQRS$ be a rectangle, where $P$ is $(a cos θ, b sin θ)$
$∴ Area of rectangle = 4 (a cos θ), (b sin θ)$
$= 2 ab sin 2 θ$
This is maximum when $sin 2 θ = 1$
Hence, maximum area $= 2 ab (1) = 2 ab$

Q. The area of the greatest rectangle that can be inscribed in the ellipse a2x2​+b2y2​=1 is

ab

2ab

ba​

ab​

Let PQRS be a rectangle, where P is (acosθ,bsinθ) ∴ Area of rectangle =4(acosθ),(bsinθ) =2absin2θ This is maximum when sin2θ=1 Hence, maximum area =2ab(1)=2ab

Q. The area of the greatest rectangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is

$2 ab$

$\frac{a}{b}$

$ab$

Let $PQRS$ be a rectangle, where $P$ is $(a cos θ, b sin θ)$
$∴ Area of rectangle = 4 (a cos θ), (b sin θ)$
$= 2 ab sin 2 θ$
This is maximum when $sin 2 θ = 1$
Hence, maximum area $= 2 ab (1) = 2 ab$