Question

Find the area of ellipse `x^2/(4) + y^2/(25)` = 1.

Answer 1

By the symmetry of the ellipse, required area of the ellipse is 4 times the area of the region OPQO.
For the region OPQO, the limits of integration are x = 0 and x = 2.

Given equation of the ellipse is `x^2/(4) + y^2/(25)` = 1

∴ `y^2/(25) = 1 - x^2/(4)`

∴ y² = `25(1 - x^2/4)`

= `(25)/(4)(4 - x^2)`

∴ y = `± (5)/(2)sqrt(4 - x^2)`

∴ y = `(5)/(2)sqrt(4 - x^2)`   ...[∵ In first quadrant, y . 0]

∴ Required area = 4(area of the region OPQO)

= `4int_0^2 y*dx`

= `4int_0^2 (5)/(2)sqrt(4 - x^2)*dx`

= `(4 xx 5)/(2) int_0^2 sqrt((2)^2 - x^2)*dx`

= `10[x/2 sqrt((2)^2 - x^2) + (2)^2/(2)sin^-1(x/2)]_0^2`

= `10{[2/2 sqrt((2)^2 - (2)^2) + (2)^2/(2)sin^-1(2/2)] - [0/2 sqrt((2)^2 - (0)^2) + (2)^2/(2)sin^-1(0/2)]}`

= 10{[0 + 2 sin^–1 (1)] – [0 + 0]}

= `10(2 xx pi/2)`
= 10π sq. units.