# reduction of $6xy + 8 y^2 -12x-26y + 11 = 0$ to canonical one of a second-order curve

I have this polynomial

$$6xy + 8 y^2 -12x-26y + 11 = 0$$

and I need to reduce it to a canonical equation of a second-order curve. The correct answer from the textbook is that it is a hyperbola

$$\frac{X^2}{1} - \frac{Y^2}{9} = 1$$

and the coordinate system must be rotated to an angle equal to

$$\arctan{3}$$

and the origin of the new coordinate system must be moved to the point $O'(-1,2)$. I checked this in Mathematica® program, so I think it is safe to assume that there are no errors in the textbook.

And now about the problem itself. I calculated the angle of rotation according to the formula in my textbook. It is (the expression above is the example of a polynom to explain indexes like $a,b,c$ etc.)

$$ax^2 + 2bxy + cy^2 + 2dx + 2ey +g = 0 \\ \cot{2\alpha} = \frac{a-c}{2b}$$

I solved this equation and got the correct answer: $\alpha = \arctan{3}$

Then I substituted $x$ and $y$ variables with their values according to the formulae:

$$x = \cos{\alpha}x' - \sin{\alpha}y'\\ y = \sin{\alpha}x' + \cos{\alpha}y'$$

Let me show this incrementally for convenience

$$6xy + 8 y^2 =\\ 6(\cos{\alpha}x' - \sin{\alpha}y')(\sin{\alpha}x' + \cos{\alpha}y') + 8(\sin{\alpha}x' + \cos{\alpha}y')^2 =\\ 6(\sin{\alpha}\cos{\alpha} * x'^2 +\cos^2{\alpha} * x'y' -\sin^2{\alpha} * x'y' - \sin{\alpha}\cos{\alpha} * y'^2) + 8(\sin^2{\alpha} * x'^2 + 2\sin{\alpha}\cos{\alpha} * x'y' + \cos^2{\alpha} * y'^2) =\\ 6\sin{\alpha}\cos{\alpha} * x'^2 + 8\sin^2{\alpha} * x'^2 - 6\sin{\alpha}\cos{\alpha} * y'^2 + 8\cos{\alpha}^2 * y'^2 + 6\cos{\alpha}^2 * x'y' -6\sin^2{\alpha} * x'y' + 16\sin{\alpha}\cos{\alpha} * x'y' = \\ (6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha})x'^2 + (- 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha})y'^2 + (6\cos{\alpha}^2 -6\sin^2{\alpha} + 16\sin{\alpha}\cos{\alpha})x'y'$$

the factor at $x'y'$ evaluates to 0 given $\alpha = \arctan{3}$, I checked this in Mathematica, so from the part above only this remains:

$$(6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha})x'^2 + (- 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha})y'^2$$

And the remaining part is:

$$-12x-26y + 11 = \\ -12(\cos{\alpha}*x' - \sin{\alpha}*y') - 26(\sin{\alpha}*x' + \cos{\alpha}*y') + 11 = \\ -12\cos{\alpha}*x' + 12\sin{\alpha}*y' -26\sin{\alpha}*x' - 26\cos{\alpha}*y' + 11 = \\ (-12\cos{\alpha} - 26\sin{\alpha})*x' + (12\sin{\alpha} - 26\cos{\alpha})*y' + 11 = \\ 2((-6\cos{\alpha} - 13\sin{\alpha}))*x' + 2(6\sin{\alpha} - 13\cos{\alpha})*y' + 11$$

So now this curve has the next equation in the rotated coordinate system (let me use $x$ and $y$ instead of $x'$ and $y'$):

$$\alpha = \arctan{3}\\ a = 6\sin{\alpha}\cos{\alpha} + 8\sin^2{\alpha} \\ c = - 6\sin{\alpha}\cos{\alpha} + 8\cos^2{\alpha} \\ d = -6\cos{\alpha} - 13\sin{\alpha} \\ e = 6\sin{\alpha} - 13\cos{\alpha}\\ g = 11\\ ax^2 + cy^2 + 2dx + 2ey + g = 0$$

Then it is stated in my textbook that one can further reduce this equation by moving the origin of a coordinate system to the point $O(-\frac{d}{a}, -\frac{e}{c})$

And this is the place where I can't get the correct answer (which is $O(-1,2)$), cause this $-\frac{d}{a}$ gives me 1.58114 instead of -1 and this $-\frac{e}{c}$ gives 1.58114 instead of 2.

I've checked this in Mathematica.

Could anyone explain me what where I'm wrong, please?

Thank you in advance.

I always try to complete the squares: $$6xy+8y^2 −12x−26y+11=0$$ $$\left[ 8y^2+2.2 \sqrt{2}y\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)\right]-12x+11=0$$ $$\left[ 8y^2+2.2 \sqrt{2}y\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)+\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)^2\right]-\left(\frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}}\right)^2-12x+11=0$$ $$\left[2\sqrt{2}y + \frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}} \right]^2-\frac{1}{8}(9x^2+18x+81)=0$$ $$\left[2\sqrt{2}y + \frac{3x}{2\sqrt{2}}-\frac{13}{2\sqrt{2}} \right]^2-\frac{1}{8}(3x+3)^2=9$$ $$\frac{1}{8}(8y+3x-13)^2-\frac{1}{8}(3x+3)^2=9$$ $$(8y+3x-13)^2-(3x+3)^2=72$$ $$(8y+3x-13)^2-\frac{(x+1)^2}{9}=72$$ $$X^2-\frac{Y^2}{9}=72$$

• thank you, I did not know about this way. I'll try it asap.
– d.k
Commented Jul 29, 2014 at 13:24
• This method can always be used to rewrite a quadratic form into a sum of squares of lineairly independent forms. It is a method invented by Lagrange. There are two rules. I used the first one (completing the squares) twice. If there are no squares left, i.e. only a term in xy, x or y, you can use the following trick to get a sum of squares: $$uv=\frac{1}{4}(u+v)^2-\frac{1}{4}(u-v)^2$$ Commented Jul 29, 2014 at 13:30
• sorry, but I missed the point. Could you please detail how can I get the right answer (that the given equation is an equation of a hyperbola, which is rotated and shifted to the point (-1,2))?
– d.k
Commented Jul 29, 2014 at 18:01
• I verified my calculations with Maple and have seen that I made an arithmetical error. I have corrected it. Commented Aug 1, 2014 at 6:46
• You can find the origin $O'(-1,2)$ by solving the equations $x'=0$ and $y'=0$, i.e. $8y+3x-13=0$ and $3x+3=0$. Commented Aug 1, 2014 at 6:50

Hint:

When $\tan \alpha = 3,\,$ then $\cos \alpha = \frac{1}{\sqrt{10}},\,\sin \alpha=\frac{3}{\sqrt{10}}\Rightarrow$

$6xy + 8 y^2 -12x-26y + 11 = 0$

$x = x'\frac{1}{\sqrt{10}} - y'\frac{3}{\sqrt{10}}\\ y = x'\frac{3}{\sqrt{10}} + y'\frac{1}{\sqrt{10}}$

$\Rightarrow 9x'^2-y'^2-9\sqrt{10}x'+\sqrt{10}y'+11=0\Rightarrow \frac{(x'-\sqrt{5/2})^2}{1}-\frac{(y'-\sqrt{5/2})^2}{9}=1$

$\Rightarrow \frac{X^2}{1}-\frac{Y^2}{9}=1$

• thank you, but the main confusing thing was why I was not able to get the correct coordinates of transformed system's origin.
– d.k
Commented Jul 30, 2014 at 10:25

Finally I have found where I was wrong.

I was getting wrong results for this expression $O(-\frac{d}{a}, -\frac{e}{c})$ - $(1.58114, 1.58114)$ instead of $(-1,2)$ because it was giving me coordinates of hyperbola's center in the already rotated coordinate system instead of the original one. To get the correct coordinates of the new system in the old system I need to perform a reverse operation, i.e. express $x,y$ of the new origin in the original system, that is perform rotation to an angle $-\alpha$ and shifting to the point $O(\frac{d}{a}, \frac{e}{c})$. Generally I need to take the new system as an "old" one.

When I did this, given $$\tan{\alpha} = 3 \\ \sin{-\alpha} = - \sin{\alpha} = - \frac{\tan{\alpha}}{\sqrt{1 + \tan^2{\alpha}}} = - \frac{3}{\sqrt{10}}, \\ \cos{-\alpha} = \cos{\alpha} = \frac{1}{\sqrt{1 + \tan^2{\alpha}}} = \frac{1}{\sqrt{10}}, \\ x'=0, \ y'=0, \ a = -\frac{d}{a} = \frac{10}{\sqrt{2}}, \ b = -\frac{e}{c} = \frac{10}{\sqrt{2}}$$

I have got a system of linear equations

$$\begin{cases} x'=a + x \cos{-\alpha} - y \sin{-\alpha} \\ y'=b + x \sin{-\alpha} - y \cos{-\alpha} \\ \end{cases} \begin{cases} 0 = \frac{10}{\sqrt{2}} + x \frac{1}{\sqrt{10}} + y \frac{3}{\sqrt{10}} \\ 0=\frac{10}{\sqrt{2}} - x \frac{3}{\sqrt{10}} + y \frac{1}{\sqrt{10}} \\ \end{cases} \implies x=1, \ y = -2$$

That is the coordinates of transformed system's origin in the old system is $O(-1,2)$, which is correct