Parabola as a locus: equal distance to a focus and a directrix
A parabola can be defined without any algebra at all: it is the set of all points \(P\) in the plane
whose distance to a fixed point (the focus) equals the perpendicular distance to a fixed line (the directrix).
This is a locus definition, meaning it describes a whole curve as a condition:
\[
PF = \text{dist}(P,\text{directrix}).
\]
Every point on the parabola satisfies this equality, and any point that satisfies the equality lies on the parabola.
To turn the definition into an equation, it is common to square both sides. If the focus is
\(F(f_x,f_y)\) and the directrix is the line \(ax + by + c = 0\), then the distance from \(P(x,y)\) to the focus is
\(\sqrt{(x-f_x)^2+(y-f_y)^2}\). The perpendicular distance from \(P\) to the line is
\[
\text{dist}(P,\text{line})=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}.
\]
Squaring removes the square root and absolute value issue (the absolute value disappears because it is squared):
\[
(x-f_x)^2+(y-f_y)^2 = \frac{(ax+by+c)^2}{a^2+b^2}.
\]
After multiplying by \(a^2+b^2\), you obtain a quadratic equation in \(x\) and \(y\). Even though the equation is quadratic,
it represents a parabola (a conic section) because it comes from a focus–directrix condition.
A key geometric parameter is \(p\), the distance from the vertex to the focus. Let \(d\) be the distance from the focus to the directrix.
Then the vertex lies halfway between the focus and the directrix along the perpendicular direction, so
\[
p=\frac{d}{2}.
\]
If the focus lies on the directrix, then \(d=0\) and \(p=0\); the parabola becomes degenerate and there is no proper “U-shaped” curve.
When the directrix is horizontal, \(y=d\), the parabola’s axis is vertical. The vertex is at
\(V\big(f_x,\frac{f_y+d}{2}\big)\) and a convenient vertex form is
\[
(x-f_x)^2 = 4p\,(y-y_V),
\]
where \(p=\frac{f_y-d}{2}\) is signed: if \(p>0\) the parabola opens upward; if \(p<0\) it opens downward.
Similarly, for a vertical directrix \(x=d\), the axis is horizontal and the vertex form is
\[
(y-f_y)^2 = 4p\,(x-x_V),
\]
with \(p=\frac{f_x-d}{2}\).
