How to turn the ellipse $x^2 - xy + y^2 - 3y - 1 = 0$ to the canonical form using an isometric transformation?

Question

There is an exam problem I'm having trouble with, it is as follows:

Turn the equation $x^2 - xy + y^2 - 3y -1 = 0$ into the canonical form using an isometric transformation and write down the transformation.

Comparing to the general equation for conic sections I see that this is a rotated ellipse. I assume that the solution would be to multiply a rotation matrix (to align the ellipse's axes to the $x$ and $y$ axes) and a translation matrix (to bring the ellipse's center to the origin), and then somehow apply the result to the ellipse.

How would I go about doing this? How do I get the coordinates of the ellipse's center and the angle of its rotation to construct the matrices, and then how would I apply the transformation to the ellipse itself?

Answer 1

To keep it understandable albeit inelegant I'll do the passage in two steps

First the translation

$x^2 - xy + y^2 - 3y -1 = 0$

we look for a new centre $(h;\;k)$ so we substitute $x=x'+h;\;y=y'+k$

$-(h+x) (k+y)+(h+x)^2+(k+y)^2-3 (k+y)-1=0$

$x'^2-x' y'+y^2+x' (2 h-k)+y' (-h+2 k-3) +h^2-h k+k^2-3 k-1=0$

To have no first degree terms we put $2h-k=0;\;-h+2 k-3=0$ which gives

$h=1;\;k=2$ and we plug these values in the previous equation

$x'^2 - x' y' + y'^2=4$

Now we want to get rid of the $x'y'$ term. To do so we have to rotate the axis using these equations

$\begin{aligned}x'&=X\cos \theta -Y\sin \theta \\y'&=X\sin \theta +Y\cos \theta \end{aligned}$

$(X \sin \theta+Y \cos \theta)^2-(X \cos \theta-Y \sin \theta) (X \sin \theta+Y \cos \theta)+(X \cos \theta-Y \sin \theta)^2=4$

collecting terms

$X^2 \left(\sin ^2\theta+\cos ^2\theta-\sin \theta \cos \theta\right)+X Y \left(\sin ^2\theta-\cos ^2\theta\right)+Y^2 \left(\sin ^2\theta+\cos ^2\theta+\sin \theta \cos \theta\right)=4$

as we want the term $XY$ off we set $\sin ^2\theta-\cos ^2\theta=0$

$\tan^2\theta=1\to \theta=\pm \dfrac{\pi}{4}$

if we want the major axis of the ellipse to be horizontal we choose $\theta=\dfrac{\pi}{4}$ and substitute this value in the last equation

$\dfrac{X^2}{8}+\dfrac{3 Y^2}{8}=1$

the equations of the roto-translation altogether are

$\begin{aligned}x&=X\cos \frac{\pi}{4}-Y\sin \frac{\pi}{4} +1\\y&=X\sin \frac{\pi}{4}+Y\cos \frac{\pi}{4} +2\end{aligned}$

or

$\begin{aligned}x&=\frac{\sqrt 2}{2}(X-Y) +1\\y&=\frac{\sqrt 2}{2}(X+Y) +2\end{aligned}$

hope this helps

Answer 2

$$\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2-3y-1=0$$ or

$$\left(x-\frac{y}{2}\right)^2+3\left(\frac{y}{2}-1\right)^2=4$$ or $$\frac{\left(x-\frac{y}{2}\right)^2}{4}+\frac{\left(\frac{y}{2}-1\right)^2}{\frac{4}{3}}=1.$$ Now, let $x'=x-\frac{y}{2}$ and $y'=\frac{y}{2}-1$...

Answer 3

You have to find an orthonormal basis of eigenvectors associated to the matrix of the quadratic part of the equation: $$Q=\begin{bmatrix} 1& -\dfrac12\\ -\dfrac12&1 \end{bmatrix}$$ The characteristic polynomial is $\;\chi_Q(\lambda)=(\lambda-1)^2-\dfrac14$, whence the eigenvalues $\lambda_1=\dfrac12$, $\;\lambda_2=\dfrac32$, and the corresponding eigenvectors $(1,1)$ and $(-1,1)$.

Normalise these vectors to obtain an orthonormal basis, and the change of basis matrix $$u=\biggl(\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad v=\biggl(-\dfrac1{\sqrt 2},\dfrac1{\sqrt 2}\biggr),\qquad P=\begin{bmatrix} \dfrac1{\sqrt2}& -\dfrac1{\sqrt2}\\ \dfrac1{\sqrt2}&\dfrac1{\sqrt2} \end{bmatrix}.$$ Usibg this matrix, we obtain the equation of the conic in the new coordinate system $(x',y')$: $$\frac12x'^2+\dfrac32y'^2-\dfrac3{\sqrt2}(x'+y')=1$$ Let's centre this equation rewriting it as \begin{align} &\phantom{\iff}\;\frac12\bigl(x'^2-3\sqrt 2x'\bigr)+\frac32\bigl(y'^2-\sqrt2 y')=1\\ &\iff\;\frac12\biggl(x'-\frac{3\sqrt 2}2\biggr)^2+\frac32\biggl(y'-\frac{\sqrt 2}2\biggr)^2=1+\frac94+\frac34=4 \end{align} So, setting \begin{cases} X=x'-\dfrac{3\sqrt2}2=\dfrac{\sqrt2}2(x+y-3),\\ Y=y'-\dfrac{\sqrt 2}2=\dfrac{\sqrt2}2(-x+y-1), \end{cases} we obtain the equation in canonical form: $$\frac{X^2}8+\frac{3Y^2}8=1.$$

Answer 4

First center the equation by definting the function $f(x,y) = x^2 - xy + y^2 - 3y - 1 = 0$ and solving the following system of equations for $(x,y)$

$$\left. \begin{aligned} \frac{\partial f(x,y)}{\partial x} & = 2x-y = 0 \\ \frac{\partial f(x,y)}{\partial y} & = -x+2y-3 = 0 \\ \end{aligned} \right\} \pmatrix{x & y} = \pmatrix{1 & 2} $$

So now we reset the coordinates to $\pmatrix{x&y} \rightarrow \pmatrix{x+1&y+2}$

The new equation is $g(x,y) = x^2-x y + y^2 -4 = 0$

Now to re-orient the conic. Use $\pmatrix{x&y} \rightarrow \pmatrix{x \cos \theta - y \sin \theta & x \sin\theta + y \cos \theta}$ and set the coefficient of $x y$ equal to zero.

$$ g = \left( 1- \frac{\sin 2 \theta}{2} \right) x^2 + \left( 1 + \frac{\sin 2 \theta}{2} \right) y^2 +2 \left( -\frac{\cos 2\theta}{2} \right) x y -4 =0 $$

$$ \left. -\frac{1}{2}\cos(2 \theta)=0 \right\} \theta =\pm \frac{\pi}{4} $$

$$\boxed{ d(x,y) = \frac{1}{2} x^2 + \frac{3}{2} y^2 -4 =0 }$$