Write the equation of a second-order line $2x^2-12xy-7y^2+8x+6y=0$ in canonical form

Question

Write the equation of a second-order line $$2x^2-12xy-7y^2+8x+6y=0$$ in canonical form.

To find its center, we have the following system of equations: $$2x-6y+4=0, \\ -6x-7y+3=0.$$ Hence, the center is at $\big(-\frac{1}{5},\frac{3}{5}\big)$. However, I don't know how to continue?

Your help is greatly appreciated

Answer 1

Recenter the curve $2x^2-12xy-7y^2+8x+6y=0$ to origin with $x\to x-\frac15$ and $y\to y+\frac35$

$$2x^2-12xy-7y^2+1=0$$

Then, rotate the axes $\theta$ degrees with $ \cos\theta= \frac1{\sqrt5}$ and $ \sin\theta= \frac2{\sqrt5}$, which corresponds to $x\to \frac1{\sqrt5}x - \frac2{\sqrt5}y$ and $x\to \frac2{\sqrt5}x +\frac1{\sqrt5}y$ to obtain the canonical equation

$$\frac{x^2}{\frac1{10}}-\frac{y^2}{\frac1{5}}=1 $$

Edit: Determine angle of rotation

Note that the equations of the asymptotes are $2x^2-12xy-7y^2=0$, with their slopes $m_1+m_2=-\frac{12}7$ and $m_1m_2=-\frac{2}7$. Then, the angle $\theta$ of the major exit can be obtained from

$$ \tan2\theta = \frac{m_1+m_2}{1-m_1m_2}=-\frac43$$

which yields $\tan\theta=2$, hence $\cos\theta= \frac1{\sqrt5}$ and $ \sin\theta= \frac2{\sqrt5}$

Answer 2

$$2x^2-12xy +8x-7y^2+6y=2(x-3y+2)^2-(3y-2)^2-7y^2+6y=$$

$$=2(x-3y+2)^2-16y^2+18y-4=2(x-3y+2)^2-16\left(y-\frac9{16}\right)^2+\frac{81}{16}-4=$$

$$=2(x-3y+2)^2-16\left(y-\frac9{16}\right)^2+\frac{17}4=0$$

Change variables and you get the equation:

$$\begin{cases}x':=x-3y+2\\{}\\ y':=y-\frac9{16}\end{cases}\;\;\implies2x'^2-16y'^2=-\frac{17}4\implies\frac{y^2}{\left(\frac{\sqrt{17}}8\right)^2}-\frac{x^2}{\left(\frac{\sqrt{17}}{\sqrt{8}}\right)^2}=1$$

So it is a hyperbola, as you can see...a vertical one, in fact ( thus the original one is a slanted vertical-ish one ).

And you are right: the center of the original hyperbola is what you wrote

Answer 3

Generally you can rotate the $x$-$y$ axis as Jan-Magnus does with the rotate transform formula from $x$-$y$ to $u$-$v$ : $$ \bigg( \begin{array}{c} u \\ v \end{array} \bigg) = \bigg( \begin{array}{cc} \cos\theta \, \sin \theta\\ -\sin \theta \, \cos \theta \end{array} \bigg) \bigg( \begin{array}{c} x \\ y \end{array} \bigg) $$ substitute $x$,$y$ with linear form of $u$,$v$,then eliminate the cross-product item $u*v$ with its coefficient equals $0$, and so on.

Alternatively you can also let $$ \begin{cases} u = a x - b y + m, \\ v = b x + a y + n. \end{cases} $$ then eliminate $u*v$ as well and extrapolate $a,b,m,n$ also get what you want.