Problem:
Given the points $$(0,1),(0,-1),(2,0),(-2,0),(1,1)$$ find the equation for the conic section that passes through these points.
My attempt:
Using the general equation for a conic section, $$ax^2+bxy+cy^2+dx+ey+f=0$$ I inserted the points to make a system of 5 equations, which was reduced to the following, with $a$ as a free variable.
$$b=-a \\c=4a \\d=e=0 \\f=-4a$$
Putting this into the general equation for a conic, I got $$ax^2-axy+4ay^2-4a=0 \ \ (1)$$
My question:
Can I determine which type of conic section this is? I read somewhere that I can determine it by finding out the sign of $b^2+4ac$, where I ended up with $b^2-4ac = -15a^2 < 0$ which is supposed to be an ellipse. However, I can't find my way back to the place where I read this. Looking at the points on a graph, it seems visually obvious that this is an ellipse. However, I can't prove it unless the mentioned method is valid.
If it really is an ellipse, how can I algebraically manipulate $(1)$ into the form most known for ellipses?