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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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  • $\begingroup$ The area of an ellipse is easily given by its major and minor axes (unlike the length of its perimeter). $\endgroup$
    – hardmath
    Nov 29, 2012 at 15:57
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    $\begingroup$ Then could you specify the major and minor axis in terms of the constants in the given equation. $\endgroup$ Nov 29, 2012 at 15:58
  • $\begingroup$ @hardmath Honestly, why doesn't SE have down-votes for comments too? $\endgroup$
    – user724085
    Feb 21, 2021 at 17:20
  • $\begingroup$ @aminabzz: Comments are considered ephemeral in the StackExchange philosophy. If you think a Comment is awful (for one reason or another), you can flag it for moderator removal. Searching on Meta Math.SE will give more details on both points. $\endgroup$
    – hardmath
    Feb 21, 2021 at 17:30

4 Answers 4

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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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    $\begingroup$ The angle of rotation to eliminate $B$ is $\frac12\tan^{-1}\left(\frac{|B|}{|A-C|}\right)$ $\endgroup$
    – robjohn
    Feb 5, 2021 at 12:36
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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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  • $\begingroup$ That is very nice! $\endgroup$
    – Lubin
    Nov 29, 2012 at 20:50
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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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  • $\begingroup$ @Dwonvoter, Please pinpoint the mistake $\endgroup$ May 11, 2014 at 16:25
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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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    $\begingroup$ This doesn't work if the axes of the ellipse are not parallel to the coordinate axes. $\endgroup$
    – TonyK
    Nov 29, 2012 at 16:17

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