The Parabola from Its Focus and Directrix

A curve defined by equal distances

Pick a point, the focus, and a line, the directrix. Now collect every point that is exactly as far from the focus as from the directrix.

Those points form a parabola.

Try it with the focus $F(0, 1)$ and the directrix $y = -1$. The point $(4, 4)$ is $5$ units from the focus and $5$ units above the directrix, so it belongs. The vertex $(0, 0)$ belongs too, $1$ unit from each.

• the parabola
• directrix y = −1

The distance to a line is measured straight across, perpendicular to it. For a horizontal directrix that is a vertical drop, so the point $(x, y)$ is $|y + 1|$ from $y = -1$.

Deriving the equation

Let $(x, y)$ be any point on the parabola. The sentence that defines it is:

$$\text{distance to } F(0, 1) = \text{distance to } y = -1$$

Write each side with the distance formula:

$$\sqrt{x^2 + (y - 1)^2} = |y + 1|$$

Both sides are distances, so neither is negative, and squaring keeps the equation equivalent:

$$x^2 + (y - 1)^2 = (y + 1)^2$$ $$x^2 + y^2 - 2y + 1 = y^2 + 2y + 1$$

The $y^2$ terms and the $1$s cancel:

$$x^2 = 4y \qquad\text{so}\qquad y = \tfrac{1}{4}x^2$$

That is a parabola, the same shape graphed in quadratic functions, now with a geometric reason behind it.

Why the squared distances leave a parabola

The distance to the focus involves both coordinates, $x^2 + (y - 1)^2$. The distance to a horizontal line involves only $y$. When the two are set equal and squared, both sides contain $y^2$ with the same coefficient, so it cancels.

What remains has $x^2$ on one side and $y$ to the first power on the other. $y$ equals a number times $x^2$, which is exactly what makes a parabola rather than a circle or some other curve.

A circle is different for that reason. It sets the distance to a point equal to a constant, so the $x^2$ and $y^2$ both survive: $x^2 + y^2 = r^2$.

The general form

Put the vertex at the origin, the focus at $(0, p)$ and the directrix at $y = -p$. The same steps give:

$$x^2 = 4py$$

Part	Where it is
vertex	$(0, 0)$, halfway between focus and directrix
focus	$(0, p)$, a distance $p$ above the vertex
directrix	$y = -p$, a distance $p$ below the vertex

A larger $p$ puts the focus farther out and makes the parabola wider. A negative $p$ puts the focus below the vertex and opens the parabola downward.

Moving the vertex to $(h, k)$ shifts everything with it:

$$(x - h)^2 = 4p(y - k)$$

Why it is called the focus

A reflector shaped like a parabola sends every ray that arrives parallel to its axis through the focus. Signals from a satellite far away arrive as parallel rays, so a satellite dish concentrates them all onto the one point where the receiver sits.

The same property runs backwards in a flashlight. A bulb at the focus sends its light off the reflector in a beam of parallel rays.

Worked examples

Common mistakes

Practice problems