# Area of an ellipse

An ellipse has equation :

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$$

Can you provide an optimum method to find it's area?

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The area of an ellipse is easily given by its major and minor axes (unlike the length of its perimeter). – hardmath Nov 29 '12 at 15:57
Then could you specify the major and minor axis in terms of the constants in the given equation. – Adwait Kumar Nov 29 '12 at 15:58

The rotation of axes can be avoided if all that is needed is the area of the ellipse.

What we do need is to translate and normalize the ellipse to be centered at the origin:

$$Ax^2 + Bxy + Cy^2 = 1$$

In this form the ellipse has area $\frac{2\pi}{\sqrt{4AC-B^2}}$. The ellipse equation can be reduced to the above form by substituting $x+h$ for $x$ and $y+k$ for $y$ and choosing $h,k$ that make the linear order terms vanish (then divide through to get constant 1 on the right side).

Details: Assuming the original equation is that of an ellipse (and not some other general conic), we replace $x$ by $x+h$ and $y$ by $y+k$, giving this:

$$Ax^2 + Bxy + Cy^2 + (2Ah+Bk+D)x + (Bh+2Ck+E)y = -(Ah^2+Bhk+Ck^2+Dh+Ek+F)$$

Solve the simple linear system for $h,k$:

$$2Ah + Bk = -D$$ $$Bh + 2Ck = -E$$

and then divide both sides by the right hand shown above to get the desired form.

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That is very nice! – Lubin Nov 29 '12 at 20:50

When rotating conics in implicit form $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0\tag{1}$$ around the origin there are 5 invariants: $$\begin{array}{rl} I_1&=A+C\\ I_2&=(A-C)^2+B^2\\ I_3&=D^2+E^2\\ I_4&=(A-C)(D^2-E^2)+2DEB\\ I_5&=F\tag{2} \end{array}$$ Assuming that we have rotated to eliminate $B$, we have $$Ax^2+Cy^2+Dx+Ey+F=0\tag{3}$$ Translating the center to the origin gives $$Ax^2+Cy^2+F-\frac{D^2}{4A}-\frac{E^2}{4C}=0\tag{4}$$ which is the same as $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{5}$$ if we set $$a=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{A}}\quad\text{and}\quad b=\sqrt{\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{C}}\tag{6}$$

Using $\text{Area}=\pi ab$ and $(4)$ and rewriting in terms of the invariants to remove the rotation, we get \begin{align} \text{Area} &=\pi\frac{\frac{D^2}{4A}+\frac{E^2}{4C}-F}{\sqrt{AC}}\\ &=\pi\frac{\color{#C00000}{2CD^2+2AE^2}-\color{#00A000}{8ACF}}{\color{#0000FF}{8(AC)^{3/2}}}\\ &=\pi\frac{\color{#C00000}{I_1I_3-I_4}-\color{#00A000}{2(I_1^2-I_2)I_5}}{\color{#0000FF}{(I_1^2-I_2)^{3/2}}}\tag{7} \end{align} Therefore, $(2)$ and $(7)$ give the area in terms of the coefficients in $(1)$.

Invariants Under Rotation and Translation

$I_1$ and $I_2$ are invariant under rotation and translation, but there is one more: the constant coefficient when the center is translated to the origin. Writing $F-\dfrac{D^2}{4A}-\dfrac{E^2}{4C}$ in terms of the rotational invariants and expanding yields $$I_6=F-\frac{AE^2-BDE+CD^2}{4AC-B^2}\tag{8}$$ Thus, the conic satisfying $(1)$, when rotated and translated becomes $$(I_1-\sqrt{I_2})x^2+(I_1+\sqrt{I_2})y^2+2I_6=0\tag{9}$$

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Using the canonical form, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, the area of an ellipse is $ab\pi$.

Hence, one way of doing it is to obtain $a,b$ from $A,\ldots,F$. This can be done by expanding the canonical form (with origin $x_0,y_0$) into your form.

$$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$$

Expand the above expression and map your polynomial coefficients to $A,\ldots,F$.

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This doesn't work if the axes of the ellipse are not parallel to the coordinate axes. – TonyK Nov 29 '12 at 16:17

As the area of a closed curve is an invariant property under the Rotation of axes,using this or this, we can remove $xy$ term to convert the given equation into $$\frac{(X-a^\prime)^2}{(A^\prime)^2}+\frac{(Y-b^\prime)^2}{(B^\prime)^2}=1$$

Consequently, the area will be $\pi A^\prime B^\prime$

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@Dwonvoter, Please pinpoint the mistake – lab bhattacharjee May 11 '14 at 16:25