Determining whether parabola is rotated, just by looking at the equation I have an equation for a parabola rotated by $45^\circ$, but let's assume I don't know it's a rotated parabola:
$x^2 + y^2 + 2xy - 8x + 8y = 0 $
I can write this as $(x+y)^2 = 8(x-y) $
But how can I tell this is the equation for a rotated parabola? And in an arbitrary case, how could I tell by how much it is rotated? Is there a systematic way of determining this?
Thanks.
 A: General equation of a parabola is $ax^2+bxy+cy^2+dx+ey+f=0$ where $b^2-4ac=0$
https://en.wikipedia.org/wiki/Rotation_of_axes#Rotation_of_conic_sections
Here you can see how to get the angle of rotation in a general case
A: As you know already, the equation rewrites $(x+y)^2 = 8(x-y).$
With 
\begin{cases}
X=x+y\\
Y=x-y
\end{cases}
we get $X^2=8Y,$ clearly an equation of parabola.
The basis vectors were transformed as follows: 
$$\begin{aligned}&(1,0)\to(1,1)\\
&(0,1)\to (1,-1)
\end{aligned}$$
The transformation is a rotation with the center $(0,0)$ and the angle $-{\pi\over4}$ composed with a scaling (the lengths are multiplied by $\sqrt 2.$)

GENERALIZATION  (see also wikipedia )
Under following constraints: 


*

*$b^2-4ac=0$

*at least one of $a,c$ is non-zero 

*the conic section given by (*) is not degenerate,


is the equation $$ax^2+bxy+cy^2+dx+ey+f=0 \quad\quad (*)$$ that of a parabola. If $b\neq 0,$ the parabola is rotated by an angle different of $k\pi / 2,\; k \in \mathbb{Z}.$
As $b^2-4ac=0$, the part $ax^2+bxy+cy^2$ is the square of a linear polynomial and can be written as $(\sqrt{|a|}x+\sqrt{|c|y})^2.$
The equation (*) takes the form similar to the above $(x+y)^2 = 8(x-y).$ 
A: The type of conic that the general equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ represents can be determined from the discriminant of its quadratic part. In particular, you have a (possibly degenerate) parabola when $$\det\begin{bmatrix}A&\frac B2\\\frac B2&C\end{bmatrix} = AC-\frac{B^2}4 = 0,$$ or more simply, when $B^2=4AC$. This also means that this matrix is singular, and it turns out that the parabola’s axis is parallel to the null space of this matrix. In your case, the associated matrix is $$\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ and it’s easy to see that its null space is spanned by $(1,-1)^T$. You’ll need to do a bit more work to determine if it opens down to the right or up to the left, though.  
One way to do this is to rewrite the equation in terms of the parabola’s principal axes. Now that you know the axis direction, you can find the parabola’s vertex in any of several ways, and from there find equations $ax+by+c=0$ and $-bx+ay+d=0$ of the axis and tangent at the vertex, respectively. The choice of signs of the coefficients in the latter equation is important here. You can then rewrite the parabola’s equation in the form $(ax+by+c)^2=k(-bx+ay+d)$. Effectively, you’ve changed to a coordinate system in which the equation has the form $Y^2=kX$. The sign of $k$ determines whether the vector $(-b,a)$ points toward the parabola’s interior (positive) or exterior (negative).  
Another possibility is to find a chord that’s perpendicular to the axis and then compute the polar of its midpoint. Since the point lies in the interior of the parabola, its polar is exterior to the parabola and perpendicular to the axis, so the direction from the polar to this point is the “positive” direction of the parabola.  
To illustrate with your parabola, setting $x=0$ and solving for $y$ yields the points $(0,0)$ and $(0,-8)$ on the parabola. The former looks very convenient, but turns out not to work for this since it’s the vertex. The perpendicular to the parabola’s axis through the latter point is $x-y+8=0$, which also intersects the parabola at $(0,8)$. The midpoint of this chord is $(4,-4)$. Its polar is $x-y-8=0$, and plugging the point into the left-hand side expression produces $16$, so the line’s normal $(1,-1)$ points toward this point, hence this is also the “positive” axis direction.
A: Given the equation of the conic as
$$
a\,x^{\,2}  + 2bxy + cy^{\,2}  + 2dx + 2ey + f = 0
$$
consider the line connecting the origin with a point on the conic, and express it in parametric form as
$$
\left\{ \matrix{
  x = \lambda \,t \hfill \cr 
  y = \mu \,t \hfill \cr}  \right.
$$
Replacing these in the conic equation, you get
$$
\left( {a\lambda ^{\,2} \, + 2b\lambda \mu \, + c\mu ^{\,2} } \right)\,t^{\,2}  + 2\left( {d\lambda  + e\mu } \right)\,t + f = 0
$$
We want to explore what the secant will become when $t \to \infty$, so we can divide first by $t^2$ and take the limit to get
$$
a\lambda ^{\,2} \, + 2b\lambda \mu \, + c\mu ^{\,2}  = 0
$$
an homogeneous quadratic equation in $\lambda , \, \mu$.
So we can solve it  by  putting  $\lambda =1$ (or $\mu = 1$) and getting a standard quadratic equation in the other variable.
As already explained in the other answers, the conic is a parabola if the resulting equation provides two coincident solutions.
Then the vector $\lambda , \, \mu$ is a vector parallel to the axis of the parabola, and directed in the same direction of the "opening" of the parabola
(the direction it moves $\to \infty$), and of course you can tell how much it is inclined.
If you know (i.e., when you will know) about homogeneous coordinates, the above will become much more .."linear"
as being the intercept of the parabola (conic) with the "line at infinity".
