# Using part of the discriminant to determine a conic?

I'm currently doing some independent study out of a book titled Algebraic Geomety: A Problem Solving Approach by Thomas Garrity. The first chapter is on conics, and is very easy up until a certain point:

$P(x,y) = ax^2 + bxy + cy^2 + dx + ey + h$

We get the discriminant by "solving for x" using the quadratic formula:

$\Delta_x(y) = (b^2-4ac)y^2 + (2bd-4ae)y + (d^2-4ah)$

Suppose $b^2 - 4ac = 0$.

(1) Show that $\Delta_x(y)$ is linear and that $\Delta_x(y) \geq 0$ if and only if $y \geq \frac{4ah-d^2}{2bd-4ae}$, provided $2bd-4ae \neq 0$.

(2) Conclude that if $b^2-4ac = 0$ (and $2bd-4ae \neq 0$), then $V(P)$ is a parabola.

The first part is very easy, but I'm surprisingly stumped on the second part. The book hasn't formally defined a parabola (or any other conics-it's a problem solving book so I think this problem stands to define these). I did find this thread: Is the discriminant of a second order equation related to the graph of $ax^2+bxy+cy^2+dx+ey+f=0$?, which mentions completing the square, which seems very useful - I'm just not sure how to show the result is a parabola.

The next two problems are essentially the same but for ellipses and hyperbolas, which seem to follow from this.

Thanks!

Note that if $y\neq 0$, $$ax^2+bxy+cy^2=y^2\left(a\left(\dfrac xy\right)^2+b\dfrac xy +c\right)$$
If $b^2-4ac=0$ and $a\neq 0$, then $at^2+bt+c=a(t-t_0)^2$ for $t_0=-\dfrac{b}{2a}$ hence $$ax^2+bxy+cy^2=ay^2\left(\dfrac xy-t_0\right)^2=a(x-t_0y)^2$$ Note that the equality $ax^2+bxy+cy^2=a(x-t_0y)^2$ is also true when $y=0$.
Thus, $P(x,y)=a(x-t_0y)^2+dx+ey+f=a(x-t_0y)^2+d(x-t_0y)+(e+dt_0)y+f$. When $e+dt_0\neq 0$, $P=0$ gives the equation of a parabola $y=a'X^2+b'X+c'$. If $e+dt_0=0$, the equation $0=a'X^2+b'X+c'$ gives two parallel lines or a single "double" line.
Finally, if $a=0$, you can exchange the roles of $x$ and $y$ in the discussion above.
Edit: Why is it a parabola? Simply because a conic $V(P)$ is a parabola if the signature of the quadratic form $ax^2+bxy+cy^2$ is $(1,0)$ or $(0,1)$, i.e. if it's a multiple of the square of a linear form.