calculusconic sections
My calculus book says
The equation $Ax^2+Bxy+Cy^2=1$ produces a hyperbola if $B^2>4AC$ and
an ellipse if $B^2<4AC$. A parabola has $B^2=4AC$.
When I set $B=2,A=C=1$, the equation becomes $x^2+2xy+y^2=1$, which is equivalent to $(x+y)^2=1$ or $y=-x\pm1$. This is not a parabola but two parallel lines. What went wrong?
This parabola cannot be written in $x=f(y)$ form (like a sideways parabola) or in $y=f(x)$ form (vertical parabola). You can see this by the mixed $xy$ term when you expand everything. It is a parabola rotated to some arbitrary angle. That's why none of your methods work. You need to first find out the rotation angle, then use a coordinate transformation to put it in a standard form and then you can do all the good things you want to do with it.
If your parabola is of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ then the angle by which the parabola has been rotated is given by $$\cot(2\theta)=\frac{A-C}{B}.$$
In your case you have $x^2+4xy+4y^2-4x+4y=0$ so $$\theta = \frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\approx-26.56^{\circ}\approx333.43^{\circ}$$
and indeed your parabola looks like
The thing here is that we actually don't need $\theta$ itself but rather we need $\sin(\theta)$ and $\cos(\theta)$ to "undo" the rotation. We can get the exact values without computing $\theta$ itself by drawing a triangle and using half-angle identities
$$\cos(\theta)=\cos\left(\frac{1}{2}\arctan(-4/3)\right)=\frac{2}{\sqrt{5}}$$ $$\sin(\theta)=\sin\left(\frac{1}{2}\arctan(-4/3)\right)=-\frac{1}{\sqrt{5}}$$
and then substitute (call this set 1)
$$x = x'\cos(\theta)-y'\sin(\theta)$$ $$y = x'\sin(\theta)+y'\cos(\theta)$$
and simplify the given parabola to get
$$-12x'+4y'+5\sqrt{5}y'^2=0.$$
So now we have a sideways parabola and notice how the mixed term $xy$ has disappeared so now you get do whatever you need to do to this rotated parabola. But after you find the directrix, vertix, focus, etc, you would have to rotate all those by $\theta$ above to get the correct values in the correct coordinate system. So we unrotate the parabola, find everything we need, and then rotate everything we found.
Continuing this example, the sideways parabola looks like this.
It is very slightly shifted and scaled. I leave the rest of algebra up to you. After you find everything you need, for some of them you will need (call this set 2)
$$x' = x\cos(\theta)+y\sin(\theta)$$ $$y' = -x\sin(\theta)+y\cos(\theta)$$
to go back to the original coordinate system.
To go from $(x',y')\rightarrow (x,y)$ use set 1. To go from $(x,y)\rightarrow (x',y')$ use set 2.
Addendum: Your method doesn't work because it only works with unrotated (canonical) parabolas meaning parabolas where the axis of symmetry is parallel to the $x$-axis or the $y$-axis. The methods you are using don't work when there is a mixed $xy$ term present so we change coordinates and eliminate the $xy$ term and then use the "shortcuts" you know. Like Blue said, you can use the easy (shortcut) methods in a complicated way or you can use complicated methods (basics/using definitions) in a simple way. For example, the definition of a parabola is that given a point (the focus) and a straight line (a directrix), a parabola is the set of all points equidistant from both. For any parabola rotated or not, you can use this definition to find the focus and the directrix. But if you want to use the
"Start with $y=ax^2+bx+c$ and put it into $4p(y-k)=(x-h)^2$ and your focus stares at you in your face with coordinates $(h,k)$ and $p$ is the distance from the vertex to the focus, and the directrix"
then your parabola MUST be a canonical form (in this case opening up or down). This shortcut for finding the focus, vertex, and directrix ONLY works if there is no $xy$ term. And that is because this shortcut is derived assuming that the parabola opens only up or down. It cannot be sideways nor at any other angle. Since your parabola isn't canonical, these shortcuts fail. So you either "make it canonical" or you use the more complicated/definition method like Blue's method. They both work. You decide which one you prefer.
There are two ways to prove this:
Formal: You can show, through a bunch of ugly computation, that the expression $B^2-4AC$ is invariant under rotation. So, consider when $B=0$ (in other words, when the conic section's directrix is parallel to one of the axes). It is easy to see that for a hyperbola $-4AC$ is positive, for an ellipse $-4AC$ is negative, and for a parabola $-4AC$ is $0$. For a better worded explanation, go to this link.
Very Informal But Intuitive: Take the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. Imagine if $x$ and $y$ were very large numbers. We can forget about $Dx+Ey+F$ because it becomes insignificant when compared to $Ax^2+Bxy+Cy^2=0$. Now, divide by $x^2$:
$$A+B\left(\frac{y}{x}\right)+C\left(\frac{y}{x}\right)^2=0$$
We notice that we now have a quadratic in $\frac{y}{x}$.
This next part is a little hard to explain in words (and my English is sort of bad) but I will try my best.
The number of solutions to this equation represents the number of ways in which the graph of the equation "zooms off towards infinity." Imagine zooming out really far from a graph of a hyperbola. You would only see an "X" formed by two lines (these lines are the asymptotes of the hyperbola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slopes of those lines. Imagine zooming out really far from a graph of a parabola. You would only see one line (the axis of symmetry for the parabola). If you solve for $\frac{y}{x}$ in the above equation, you would be solving for the slope of that line. If you zoomed out really far from a graph of an ellipse, you would see a point.
So, if $A+B\left(\frac{y}{x}\right)+C\left(\frac{y}{x}\right)^2=0$ has two solutions for $\frac{y}{x}$, the equation is a hyperbola. One solution means parabola. Zero solutions means ellipse or circle. The number of solutions corresponds to the sign of $B^2-4AC$.
I sort of like this informal proof because it explains why the discriminant of a conic looks like that of a quadratic.
Best Answer
You are right: a nondegenerate parabola doesn't admit an equation of the form $Ax^2+2Bxy+Cy^2=1$.
Indeed, if $B^2=AC\ne0$, we can multiply all coefficients by $A$, finding $$ A^2x^2+2ABxy+B^2y^2=B $$ that becomes $(Ax+By)^2=B$. If $B>0$ this factors as $$ (Ax+By-\sqrt{B})(Ax+By+\sqrt{B})=0 $$ which is a pair of parallel lines.
If $B<0$, the factorization is over the complex numbers and we get no real point.
If $B=0$, then either $A=0$ or $C=0$, so the equation becomes either $Cy^2=1$ or $Ax^2=1$, again a situation as before.
Only nondegenerate conics with a center admit an equation of the given form, which is obtained by translating the center of the conic in the origin.