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Formulas/maths/Ellipse/Area of an Ellipse

Area of an Ellipse

The ellipse x²/a² + y²/b² = 1 encloses area πab. Reduces to πa² for a circle (b = a). Derived by scaling the unit circle uniformly in one direction.
Derivation

Method 1 — Scaling argument:

The ellipse x2/a2+y2/b2=1x^2/a^2 + y^2/b^2 = 1 is obtained from the circle x2+y2=a2x^2 + y^2 = a^2 by the transformation (x,y)(x,by/a)(x, y) \mapsto (x, by/a), which scales all yy-coordinates by b/ab/a.

A uniform scaling in one direction scales all areas by the same factor. Therefore:

Area of ellipse=ba×Area of auxiliary circle=ba×πa2=πab\text{Area of ellipse} = \frac{b}{a} \times \text{Area of auxiliary circle} = \frac{b}{a} \times \pi a^2 = \pi ab

Method 2 — Integration:

For the upper half of the ellipse, y=b1x2/a2y = b\sqrt{1 - x^2/a^2}. Total area:

A=2aab1x2a2dxA = 2\int_{-a}^{a} b\sqrt{1-\frac{x^2}{a^2}}\,dx

Substitute x=asinθx = a\sin\theta, dx=acosθdθdx = a\cos\theta\,d\theta:

A=2π/2π/2bcosθacosθdθ=2abπ/2π/2cos2θdθA = 2\int_{-\pi/2}^{\pi/2} b\cos\theta \cdot a\cos\theta\,d\theta = 2ab\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta =2abπ2=πab= 2ab \cdot \frac{\pi}{2} = \pi ab

Special cases: a=b=ra = b = r gives πr2\pi r^2 (circle); b0b \to 0 gives area 0\to 0 (degenerate ellipse collapses to a line segment).