Classification of a quadric

Question

Is it possible to classify the following quadric along with its centre without using the principal axis theorem?

$$x^2 = 1 + x +xy$$

For instance, a simple equation like $x^2 + 2x = 3$ turns out to be a line with a centre at $x = -1$ by the use of completing the square and simple a simple computation. I haven't found a similar way for the equation given above.

greetings Hofmusicus

Answer 1

This solution is a more general staff about rotation quadric. In order to get the quadric at the form $(x/a)^2-(y/b)^2=1$ we have to rotate the axis because the original is not parallel to the symmetry axis of the quadric. Let's choose the new pair of axis $x_1$ and $y_1$, so we want to find an angle $\theta$ where we can write the quadric of the form $(x_1/a)^2-(y_1/b)^2=1$. $$\begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \\ \end{pmatrix} \begin{pmatrix} x_1\\ y_1 \end{pmatrix}= \begin{pmatrix} x\\ y \end{pmatrix} $$ That give us $$x=x_1cos\theta-y_1sin\theta$$ $$y=x_1sin\theta+y_1cos\theta$$ Replacing both equations at the original we get $$(x_1cos\theta-y_1sin\theta)^2=1+(x_1cos\theta-y_1sin\theta)+(x_1cos\theta-y_1sin\theta)(x_1sin\theta+y_1cos\theta)(*)$$ But considering that in the new pair of axis we don't want the term $x_1y_1$ so we choose $\theta$ properly in order to that occur. So the coefficient of $x_1y_1$ must be zero. $$-2sin\theta\cos\theta=cos^2\theta-sin^2\theta \Rightarrow -sin2\theta=cos2\theta \Rightarrow tg2\theta=-1$$ and then we can choose $2\theta=3\pi/4\Rightarrow\theta=3\pi/8$ and then $sin\theta=(\sqrt{2+\sqrt{2}})/2$ and $cos\theta=(\sqrt{2-\sqrt{2}})/2$. Considering those values you can back to expression $(*)$ and completing square get a expression of the form: $$\left(\frac{x_1}{a}\right)^2-\left(\frac{y_1}{b}\right)^2=1$$ I hope that helps.

Answer 2

we get by solving for $y$: $$x^2-1=x(1+y)$$ $$y=\frac{x^2-x-1}{x}$$ if $x$ is not zero.

Answer 3

What you have here is a Rotation of Axes. What you want to do is get rid of the $xy$ term using the method in the link, find out how much the function is rotated (which you need to do to get rid of the $xy$ term anyway), find the vertex of that function, and then rotate it back to get a point on the axis, and then use the angle of rotation to find the slope of the line that goes through that point.

Answer 4

In case of hyperbola, we may find the asymptotes instead

Rewriting as $$x = \frac{x^2-x-1}{y}$$

Take $y\to \infty$, $$x=0$$

Rewriting as $$x-y = 1+\frac{1}{x}$$

Take $x\to \infty$, $$x-y=1$$