Convert this ellipse equation, $2x^2 + 6xy + 5y^2 = 1$ into standard form.

Question

I have tried this first I took $x^2 + y^2 + 2xy$ on side and on the other side $x^2 + 4y^2 + 4xy$ and I complete square on both and this becomes $(x+y)^2 + (x + 2y)^2 = 1$ but this two line are not perpendicular to each other.

Answer 1

Change the equation of an ellipse given in general form to what you say standard form, is frequently something tedious and still more tedious when the rectangular term is not null as in the case of the equation $$2x^2+6xy+5y^2=1$$ We show here a solution to motivated beginners.

A method is to find the angle $\theta$ in $\tan(2\theta)=\dfrac{6}{2-5}=-2$ (corresponding to $\dfrac{b}{a-c}$ in the general equation $ax^2+bxy+cy^2+dx+ey+f=0$) which gives $\theta\approx-0.553574358897$ and change of coordinates from $(x,y)$ to $(x',y')$ vía $$x=x'\cos\theta-y'\sin \theta\\y=x'\sin\theta+y'\cos\theta$$ which gives, (by confort we use $(x,y)$ instead of $(x',y')$) $$(2\cos^2\theta+5\sin^2\theta+6\sin\theta\cos\theta)x^2+6(\sin\theta\cos\theta+\cos^2\theta-6\sin^2\theta)xy+(2\sin^2\theta+5\cos^\theta-6\sin\theta\cos\theta)y^2=1$$ Calculation gives $$(0.14589803375)x^2+(6.85410196625)y^2=1$$ so we have the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ where $a=2.61803398875$ and $b=0.38196601125$.

In the attached figure the black ellipse is the given one and the red ellipse is the new, after the changement of coordinates.

Answer 2

That is standard form that you got for the change of variables $(x+y,x+2y).$ In general, when faced with this kind of problem; ie, turn $$ax^2+bxy+cy^2=d$$ to standard form, just put $$x=\frac{p+q}{\sqrt{a}},\ y=\frac{p-q}{\sqrt{c}}.$$ This will always make the $xy$ term $\frac{b(p^2-q^2)}{\sqrt{ac}}.$ When you solve for $(p,q)$ in terms of $(x,y)$, you can always use matrices. This will always give orthogonal axes easily and algebraically.

---

So try the substitution $(x,y)=\left(\frac{p+q}{\sqrt{2}},\frac{p-q}{\sqrt{5}}\right).$

Substituting the above, we get

$$2(p^2+q^2)+\frac{6}{\sqrt{10}}(p^2-q^2)=1,$$ whereby we obtain $$\left(\frac{20+6\sqrt{10}}{10}\right)p^2+\left(\frac{20-6\sqrt{10}}{10}\right)q^2=1.$$ Now, this is your standard form. This is generally how you eliminate the $xy$ term without difficulty. Solving for $(x,y)$ in terms of $(p,q)$;

$$p+q=\sqrt{2}x\\ p-q=\sqrt{5}y\\ p=\frac{\sqrt{2}x+\sqrt{5}y}{2},\ q= \frac{\sqrt{2}x-\sqrt{5}y}{2}.$$