How to bring elliptical equation $2x^2+2y^2+3xy-x-y=0$ into canonical form

Question

I have this ellipse: $$2x^2+2y^2+3xy-x-y=0$$

Canonical form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

How can I bring my ellipse to that canonical form? It seems like I need some substitution.

Answer 1

I tried to look for a good way to substitute something but failed miserably ...

So instead I tried the following :

Firstly I found the centre of the ellipse by solving the equations :

$$\frac{\delta f}{\delta x} = 0 \ and \ \frac{\delta f}{\delta y} = 0$$

So the centre came out to be ($\frac{1}{7}$,$\frac{1}{7}$)

The major and minor axes will pass through this point so let the slope of the line with acute angle to the x-axis be m. (m>0)

One of the axes : $(y-\frac{1}{7})=m(x-\frac{1}{7})$

Subsequently the other axis is : $(y-\frac{1}{7})=-\frac{1}{m}(x-\frac{1}{7})$

Rearrange these equations a bit

Now the canonical form of ellipse can be rewritten as :

$$\frac{(distance \ from \ an \ axis )^2}{a^2} +\frac{(distance \ from \ other \ axis )^2}{b^2} = 1$$

The distance of a point(x,y) from a line $y=mx +c$ is given by $\frac{|y-mx-c|}{\sqrt{m^2+1}}$

So write the equations of assumed axes as $y=mx+c$ and fill them in the above equation

$$\frac{(y-mx-\frac{1}{7}+\frac{m}{7})^2}{(1+m^2)a^2}+\frac{(my+x+\frac{1}{7}-\frac{m}{7})^2}{(1+m^2)b^2}=1$$

Now simply compare coefficients of obtained equation and given equation. The coefficients of x and y are equal in given equation, so let us just do that in our equation.

$$\frac{m^2}{(1+m^2)a^2} + \frac{1}{(1+m^2)b^2} = \frac{1}{(1+m^2)a^2} + \frac{m^2}{(1+m^2)b^2}$$

$$\frac{m^2 -1}{(1+m^2)a^2} = \frac{m^2 -1}{(1+m^2)b^2}$$

So either $m=\pm 1$ or $a=\pm b$

We can eaisilly discard the possibility of $a=\pm b$ as the would mean it is a circle, but in the given equation coefficient of $xy$ is 3 (which must be $0$ for it to be a circle)

Also notice how for m=1 , the equation will have non-zero constant term; so only possible value is 1

Now plug in the value of m as -1 and again compare coefficients

Compare coefficients of $xy$:

$$-\frac{1}{b^2}+\frac{1}{a^2} = 3..........(1)$$ Compare coefficients of $x^2$:

$$\frac{1}{a^2}+\frac{1}{b^2}=4..........(2)$$

solve (1) and (2) to get $\frac{1}{a^2}$ and $\frac{1}{b^2}$;plug in the obtained values in the equation to get the required equation in canonical form.

$$\frac{7(y+x-\frac{2}{7})^2}{4} +\frac{(x-y+\frac{2}{7})^2}{4} = 1$$

Answer 2

Rewrite the equation,

$$\begin{align} & 2x^2+2y^2+3xy-x-y\\ & = 2(x+y)^2 -xy -(x+y)\\ & = 2(x+y)^2 -\frac14[(x+y)^2-(x-y)^2] -(x+y)\\ & =\frac74 \left(x+y-\frac27\right)^2+\frac14 (x-y)^2 - \frac17=0 \\ \end{align}$$

Then, let $u=\frac 1{\sqrt2}(x+y-\frac27)$ and $u=\frac 1{\sqrt2}(x-y)$ to get the canonical form

$$\frac{u^2}{\left(\frac{\sqrt2}7\right)^2} + \frac{v^2}{\left(\sqrt{\frac{2}7}\right)^2} =1$$

Answer 3

The center of the ellipse is where the gradient vanishes,

$$4x+3y-1=0,\\4y+3x-1=0.$$

Then we center around the solution $\left(\dfrac17,\dfrac17\right)$,

$$2\left(x+\frac17\right)^2+2\left(y+\frac17\right)^2+3\left(x+\frac17\right)\left(y+\frac17\right)-\left(x+\frac17\right)-\left(y+\frac17\right) \\=2x^2+3xy+2y^2-\frac17=0.$$

In matrix form,

$$z^TAz-\frac17=0$$ where $A=\begin{pmatrix}2&\frac32\\\frac32&2\end{pmatrix}$.

We diagonalize the matrix and find the Eigenvalues $\dfrac12$ and $\dfrac72$. Hence the reduced equation

$$\frac{x^2}2+\frac{7y^2}2=\frac17.$$

Answer 4

Hint. To eliminate the $xy$ term, rotate the axes by $45°.$ This gives the transformation equations $$x=\frac{u}{\sqrt 2}-\frac{v}{\sqrt 2},\,y=\frac{u}{\sqrt 2}+\frac{v}{\sqrt 2}.$$ Now you have an equation involving only quadratic terms that are squares. Then complete the squares and make a last substitution. Behold your ellipse in standard form!