Question

find the are of ellips x^2/a^2+y^2/b^2=1

Answer 1

﷮2﷯﷮ ﷮2﷯﷯+ ﷮2﷯﷮ BA is in 1st Quadrant ∴ = ﷮﷯ ﷮ ﷮2﷯− ﷮2﷯﷯ Area of ellipse = 4 × 0﷮﷮.﷯ = 4 0﷮﷮ ﷮﷯﷯ ﷮ ﷮2﷯− ﷮2﷯﷯ = 4﷮﷯ 0﷮﷮ ﷮ ﷮2﷯− ﷮2﷯﷯ ﷯ = 4﷮﷯ ﷮2﷯ ﷮ ﷮2﷯− ﷮2﷯﷯+ ﷮2﷯﷮2﷯ sin﷮−1﷯﷮ ﷮﷯﷯﷯﷮0﷮﷯ = 4﷮﷯ ﷮2﷯ ﷮ ﷮2﷯− ﷮2﷯﷯+ ﷮2﷯﷮2﷯ sin﷮−1﷯﷮ ﷮﷯− 0﷮2﷯ ﷮ ﷮2﷯−0﷯− ﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 0﷯﷯﷯﷯ = 4﷮﷯ 0+ ﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 1﷯−0−0﷯﷯ = 4﷮﷯ × ﷮2﷯﷮﷯ sin﷮−1﷯﷮ 1﷯﷯ = 2 × sin﷮−1﷯﷮ 1﷯﷯ = 2 × ﷮2﷯ = ∴ Required Area = square units﷮2﷯﷯

Answer 2

Answer:

Step-by-step explanation:

This is the general equation of an ellipse whose area is (pi)ab.

Explanation : Using Calculus :

Solve the above equation for y , y = + or - b √ [ 1 - x^2/ a^2 ]

Now use integrals to find the area of the upper right quarter of the ellipse as follows

(1 / 4) Area of ellipse = integrate(b √ [ 1 - x^2/ a^2] dx from 0 to a

Evaluate the integral, (1 / 4) Area of ellipse = (1/2) b a [ (1/2) sin(2t) + t ] for t = 0 to pi/2

Value = (1/4)pi(ab)

area of full ellipse = Value * 4 = pi(ab)