Deriving the Standard Equation y² = 4ax

The locus definition from the previous section gives an equation for the parabola no matter how the focus and directrix happen to sit in the plane, but that equation looks messy unless the axes are chosen well. The standard trick is to place the origin exactly halfway between the focus and the directrix, with the axis of the parabola along one of the coordinate axes — this single choice is what turns a general second-degree mess into the clean equation $y^2=4ax$.

Let the focus be $S$ and the directrix be the line $l$. Drop a perpendicular from $S$ to $l$, meeting it at $Z$, and let $A$ be the midpoint of $SZ$. Because $A$ is equidistant from $S$ and from $l$ (it sits exactly on $l$'s perpendicular through $S$, at half the $SZ$ distance), $A$ itself lies on the parabola — it is called the vertex. Now set up coordinates with $A$ at the origin, the axis $SZ$ along the $x$-axis, and the line through $A$ parallel to the directrix as the $y$-axis. If the distance $AS = a$ (with $a>0$), then $S=(a,0)$ and $Z=(-a,0)$, so the directrix is the vertical line $x=-a$, i.e. $x+a=0$.

Now take any point $P(x,y)$ on the locus, and let $M$ be the foot of the perpendicular from $P$ to the directrix; since the directrix is $x=-a$, we have $PM = x+a$. The defining condition $SP=PM$ (since $e=1$) becomes, after squaring both sides to remove the distance formula's square roots,

Expanding both sides and cancelling the common $x^2$ and $a^2$ terms leaves

It is worth checking the converse too: if $(x,y)$ satisfies $y^2=4ax$, then $SP=\sqrt{(x-a)^2+y^2}=\sqrt{(x-a)^2+4ax}=\sqrt{(x+a)^2}=|x+a|=PM$, so every point satisfying the equation genuinely lies on the locus. This confirms $y^2=4ax$ is the parabola, not just a curve that contains it.

A few features of this equation are worth reading off directly, since they recur constantly: setting $y=0$ gives $x=0$, so the curve passes through the origin (the vertex); setting $x=0$ gives $y=0$ (twice), so the $y$-axis touches the curve only at the vertex; and since $y^2=4ax\ge 0$ requires $x\ge 0$ (given $a>0$), the whole curve lies in the region $x\ge 0$, opening to the right, symmetric about the $x$-axis (because $y=\pm\sqrt{4ax}$ always come in a pair). As $x\to\infty$, $|y|\to\infty$ too, so the curve is an open, unbounded arc, not a closed shape like an ellipse.

Worked example. Find the focus and directrix of the parabola $y^2=20x$, and verify the point $(5,10)$ lies on it. …