Let $A;B;C;D;E$ be real numbers such that $A$ is positive and $C$ is negative. The set of points (x; y) that satisfy $Ax^{2} + Bx + Cy^{2} + Dy + E = 0 $ is (usually) a hyperbola.
a) Show this
b) Why "usually"? Can this set be something other than a hyperbola? (Hint: Complete the square; for example, if $A \neq 0$ write $Ax^{2} + Bx$ as $A(x + \frac{B}{2A})^{2}$ + a constant.
Attempts:
With regards to a) I think I may have indirectly worked through b). I didn't get a final solution nonetheless, but after working on a) and then glancing at b) this may have been the case. Anyways
a) this asks to show that the above expression is usually a hyperbola based on the given conditions. So to me that meant I had to try and put the equation into the following form"
$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$$
As this isone of the general forms of the hyperbola. That being said I looked at the expression given and I actually completed the square with respects to $x$ but I also did it with respect to $y$. Here are the crucial steps, given that $C$ is supposed to be seen as negative I labelled it $-C$:
$$ Ax^{2} + Bx - Cy^{2} + Dy + E = 0 \\ (Ax^{2} + Bx) - (Cy^{2} + Dy) + E = 0 \\ \bigg(x + \frac{B}{2A}\bigg)^{2} - \frac{C}{A}y^{2} + \frac{D}{A}y + \frac{E}{A} = \frac{B^{2}}{4A^{2}} \\ \text{now completeing the square of the y term}:\\ \frac{A}{C}\bigg(x +\frac{B}{2A}\bigg)^{2} - \bigg(y - \frac{D}{2c}\bigg)^{2} = \frac{B^{2}}{4AC} - \frac{D^{2}}{4c^{2}} - \frac{E}{C} $$
And this is where I'm stuck. If I divide everything by the constant that is on the right side of the equation I'll at least have the $1$ on the right side, but the left side won't be exactly in the form that I'm attempting to arrive at.
Questions: Did I even do the right thing for 1) and how should I approach 2)?
The equation you gave restricts the center of the hyperbola to the origin. It would be more correct to say that the (one of) the general equations is $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =1$$ Dividing by the term on the right should be sufficient from there.
For part (b), as suggested by @John Omielan, notice that if the RHS is zero then the equation reduces to
$$\sqrt{\frac AC} \left(x + \frac{B}{2A}\right) = \pm \left(y -\frac{D}{2C} \right) $$ which is clearly not a hyperbola, but a set of two lines.