Can every parabola be written in the form of a quadratic $y=ax^2+bx+c$ or $x=dy^2+ey+f$?

Question

I understand that the graph of any equation of the form $y=ax^2+bx+c$ is a parabola (please correct me if I am mistaken).

My question is about the converse: Can every parabola be written in the form of an equation $y=ax^2+bx+c$ or $x=dy^2+ey+f$?

Answer 1

To sum up the comments: No.

But you're not completely wrong. Parabolas are "just" geometric shapes, they don't require a coordinate system, but if you add a coordinate system with axes parallel to the directrix and the axis if symmetry (see Wikipedia in parabolas) if can be described by one of the two forms you have listed.

Answer 2

A parabola is just the locus of points (a collection) that is equidistant from a point and a straight line. Let $(x,y)$ be the set of points, then for let's say you have a fixed point $(p,q)$ and a line $ax+by+c=0$,

The distance between $(x,y)$ and the straight line is given by:

$$\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$$

We want to set that to be equal to the distance between $(p,q)$ and $(x,y)$

$$ \frac{|ax+by+c|}{\sqrt{a^2+b^2}}=\sqrt{(x-p)^2+(y-q)^2}$$

$$\frac{a^2x^2+b^2y^2+2abxy+2acx+2bcy}{a^2+b^2}=x^2-2px+p^2+y^2-2qx+q^2$$

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$$b^2x^2+a^2y^2-(2p(a^2+b^2)+2ac)x-(2q(a^2+b^2)+2bc)y-2abxy+(p^2+q^2)(a^2+b^2)=0$$

So the formula doesn't look pleasant, but we know from that, any parabola will be in the form of $ax^2+by^2+cx+dy+exy+f=0$

Answer 3

If it is any parabola in the plane you would need $$ax^2+by^2+cxy+dy+ex+f = 0$$, this becomes evident if you try and for example rotate your parabola: $$x_0 = \cos(\alpha)x + \sin(\alpha)y\\y_0 = -\sin(\alpha)x + \cos(\alpha)y$$ You will be getting the "mixed terms" containing $xy$ when converting $x^2$ and $y^2$ into $x_0$ and $y_0$. Some special $\alpha$ maybe pose no problem but for most of them you will be getting the cross-terms.

Answer 4

The parabolas whose axes of symmetry is parallel to x- or y- axis belong to the category you mention. They are characterized by an absence of $xy$ term which when present, makes its axis of symmetry inclined to the coordinate axes.

From the general equation of conics using an invariance condition $I_2=0$ for unity eccentricity of parabola we have

$$ a x^2 + 2 h x y + b y^2 + 2 g x+ 2 f y + 1 = 0,\, h^2 - a b =0 ;$$

when solved for $y$ as a quadratic and constants are re-lumped, it assumes a form for the inclined parabola:

$$ \boxed {y = (A x + B) \pm \sqrt{ C x + D } } $$

The two branches are separated by vertical/horizontal tangents when $ C x + D =0 $