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Prove that the area of the ellipse $$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0,$$ where $AC-B^2>0$, is equal to $$S=\frac{- \pi \Delta}{(AC-B^2)^{3/2}},$$ where $$\Delta =\begin{vmatrix}A&B&D\\B&C&E\\D&E&F\end{vmatrix}.$$

I could prove that area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\pi a b$ because I know its vertices, center, etc., but in this question I do not know the vertices of the ellipse, the major axis equation, the minor axis equation, the center, etc. So I could not solve it. Can someone assist me in this question?

Brahmagupta
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  • You have a rotation. Why don't you first of all try to eliminate that $2Bxy$? – Tolaso Aug 05 '15 at 14:20
  • More generally, one can attempt to describe a series of transformations that put a general ellipse in the normal form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ described in the question and then show that $S$ is invariant under each of these transformations. – Travis Willse Aug 05 '15 at 14:56
  • My answer here may be able to help connect the dots: http://math.stackexchange.com/a/1127581/97045 – DanielV Aug 05 '15 at 15:09

2 Answers2

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Assume WLOG that $A>0$ (otherwise multiply the equation by $-1$). We have $$ \left[\matrix{x\\ y}\right]^T\left[\matrix{A & B\\ B & C}\right]\left[\matrix{x\\ y}\right]+2\left[\matrix{D\\ E}\right]^T\left[\matrix{x\\ y}\right]+F=0. $$ After completing the square we get $$ \left(\left[\matrix{x\\ y}\right]+\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]\right)^T\left[\matrix{A & B\\ B & C}\right]\underbrace{\left(\left[\matrix{x\\ y}\right]+\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]\right)}_{=z}=\\ =\underbrace{\left[\matrix{D\\ E}\right]^T\left[\matrix{A & B\\ B & C}\right]^{-1}\left[\matrix{D\\ E}\right]-F}_{=\alpha^2}. $$ If we factorize the matrix of the quadratic form (which is positive definite) $$ M=\left[\matrix{A & B\\ B & C}\right]^{1/2} $$ the ellipse equation becomes $$ \left\|\frac{1}{\alpha}Mz\right\|=1 $$ which means that the image of the ellipse under the linear mapping $y=\frac{1}{\alpha}Mz$ is the unit circle with area $\pi$. Area deformation under linear/affine maps is the determinant, i.e. the area of the ellipse $S$ satisfies $$ \pi=\det(\frac{1}{\alpha}M)S=\frac{1}{\alpha^2}\det(M)S\qquad\Rightarrow\qquad S=\frac{\pi\alpha^2}{\det(M)}=\frac{\pi\alpha^2}{\sqrt{AC-B^2}}. $$ Finally, to calculate $\alpha^2$ we use the Schur determinant formula $$ \Delta=\det\left[\matrix{A & B & D\\ B & C & E\\D & E & F}\right]= \det\left[\matrix{A & B\\ B & C}\right]\cdot (-\alpha^2)=-(AC-B^2)\alpha^2 $$ which gives $$ S=\frac{-\pi\Delta}{(AC-B^2)^{3/2}}. $$

A.Γ.
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  • I've never seen someone complete the square with matrices like that before. Thanks for that. – DanielV Aug 05 '15 at 15:52
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    @DanielV Glad that it helps. It is the same idea as long as the matrix $H=H^T$ is invertible to complete $x^THx+2b^Tx=(x+H^{-1}b)^TH(x+H^{-1}b)-b^TH^{-1}b$, often useful when working with quadratic forms. – A.Γ. Aug 05 '15 at 16:10
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Read up about three invariants to gain a broader understanding:

$$ I_1= A + B $$ $$ I_2 = (A C - B^2) $$ $$ I_3 = \Delta $$

Several other geometrical properties are also linked to these invariants.

Narasimham
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  • Those invariants would have to be under some normalization of the conic equation, such as when $F=1$ or when $F=0$ . – DanielV Aug 05 '15 at 15:15