The ellipse $E:=\lbrace(x,y) \in \mathbb{R^2}:x^2+y^2+xy-x-y \leq 1\rbrace$ is given.

How to compute the area/Lebesgue measure of this set? Also, how an affine transformation $F: \mathbb{R^2} \to \mathbb{R^2}$ can be found such that $F(\lbrace (x,y) \in \mathbb{R^2}:x^2+y^2 \leq1 \rbrace)=E$?

I know that the area of an ellipse is $\pi \cdot $A$\cdot$B, but in this case the values of A, B aren't given.

To find an affine transformation I tried to use that such transformations on $\mathbb{R^2}$ are given by $T((x,y)^T)= A \cdot (x,y)^T+z$ for $A \in \mathbb{R^{2x2}}, z \in \mathbb{R^2}$.

If $(0,0)$ maps to the center, $(1,0)$ and $(0,1)$ to the right vertex and top co-vortex, I get with the formula:

$T((0,0)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (0,0)^T+z=z$

$T((1,0)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (1,0)^T+z=(a,c)^T+z$

$T((0,1)^T)=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot (0,1)^T+z=(b,a)^T+z$

This looks wrong to me, since I should get a linear system with six equations which can be solved.

So what exactly has to be done to determine the area of this ellipse and find the affine transformation, that maps the unit circle to this ellipse?

  • If $z\in\mathbb R^2$ as it appears to be, then after equating these three expressions to the center and vertices, you have the six equations that you’re expecting. – amd Nov 9 at 1:59
up vote 1 down vote accepted

$x^TAx + Bx = 1$

$x^2 + xy + y^2 - x - y = 1\\\pmatrix{x & y} \pmatrix { 1 &\frac 12 \\\frac 12 & 1} \pmatrix {x\\y} + \pmatrix{-1&-1} \pmatrix {x\\y} = 1$

Diagonalize the matrix on the left.

$\pmatrix { 1 &\frac 12 \\\frac 12 & 1} = \pmatrix{\frac {\sqrt {2}}{2}&\frac {\sqrt {2}}{2}\\-\frac {\sqrt {2}}{2}&\frac {\sqrt {2}}{2}} \pmatrix { \frac 12 &0 \\ 0& \frac 32}\pmatrix{\frac {\sqrt {2}}{2}&-\frac {\sqrt {2}}{2}\\\frac {\sqrt {2}}{2}&\frac {\sqrt {2}}{2}} $

let $\pmatrix {x'\\y'} = \pmatrix{\frac {\sqrt {2}}{2}&-\frac {\sqrt {2}}{2}\\\frac {\sqrt {2}}{2}&\frac {\sqrt {2}}{2}}\pmatrix {x\\y}$

$x^2 + xy + y^2 - x - y = 1\\ \pmatrix{x'&y'}\pmatrix {\frac 12 \\ &\frac 32}\pmatrix{x'\\y'} - \sqrt 2 y' = 1$

$\frac 12 x'^2 + \frac 32 y'^2 - \sqrt 2 y' = 1\\ \frac 12 x'^2 + \frac 32 (y'^2 - \frac {2\sqrt 2}{3} y' + \frac {2}{9}) = 1+\frac 13\\ \frac {x'^2}{\frac 83} + \frac {y'^2}{\frac 89} = 1\\ $

$A = \pi\sqrt {\frac 83 \cdot \frac 89} = \pi \frac 89\sqrt 3$

You get rid of the $xy$ term by an orthogonal transformation, $$ x = \frac{u+v}{\sqrt 2} \; , \; $$ $$ y = \frac{u-v}{\sqrt 2} \; . \; $$

The result is an evident ellipse with center not at the $u,v$ original, so you still need to complete some squares to find out the constant term.

When finished, you have $a(u-u_0)^2 + b (v - v_0)^2 = c$ which has the same area as $a u^2 + b v^2 = c$

Hint. If we rotate by $\pi/4$ the area does not change. This done by repalcing $x$ and $y$ according to $$ x\longrightarrow\frac{x+y}{\sqrt{2}}, \quad y\longrightarrow\frac{x-y}{\sqrt{2}} $$ and obtain $$ \left(\frac{x+y}{\sqrt{2}}\right)^2+\left(\frac{x-y}{\sqrt{2}}\right)^2+ \left(\frac{x+y}{\sqrt{2}}\right)^2\left(\frac{x-y}{\sqrt{2}}\right)^2-\left(\frac{x+y}{\sqrt{2}}\right)-\left(\frac{x-y}{\sqrt{2}}\right)\le 1, $$ or $$ \frac{3}{2}x^2+\frac{1}{2}y^2-\sqrt{2}x\le 1, $$ or $$ \frac{3}{2}\left(x^2-\frac{2\sqrt{2}}{3}x+\frac{2}{9}\right)+\frac{1}{2}y^2\le \frac{4}{3}, $$ or $$ \frac{(x-\frac{\sqrt{2}}{3})^2}{\left(\frac{2\sqrt{2}}{3}\right)^2}+\frac{y^2}{\left(\frac{2\sqrt{2}}{\sqrt3}\right)^2}\le 1. $$

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