Parabolas Advanced. This lesson goes beyond core Algebra II. You can skip it.

The distance condition that defines a parabola

Try the numbers first. Put a point $F = (0,1)$ and, two units below it, a horizontal line $d\colon y = -1$. Measure two points against both of them.

• $(0,0)$ is $1$ unit from $F$ (straight up) and $1$ unit from $d$ (straight down). The two distances match.
• $(4,4)$ is $\sqrt{4^2+3^2}=5$ units from $F$, and $4-(-1)=5$ units from $d$. They match again.
• $(4,3)$ is $\sqrt{4^2+2^2}=\sqrt{20}\approx 4.47$ units from $F$, but only $3-(-1)=4$ units from $d$. Here the two distances disagree.

Collect every point whose distance to $F$ equals its distance to $d$, not just the two that matched above, and the result is a curve: a parabola. The point $F$ is its focus, and the line $d$ is its directrix.

Written as a general rule instead of two specific points: fix a point $F$ and a line $d$ that does not pass through $F$. The parabola with focus $F$ and directrix $d$ is the set of all points $P$ in the plane satisfying

that is, every point whose distance to $F$ equals its distance to $d$. A curve defined this way, by a condition every point must pass rather than by a formula, is called a locus.

The condition that $F$ is not on $d$ matters. If you put the focus on the directrix, the only points that stay equidistant are the ones on the perpendicular to $d$ through $F$. In that case the “curve” collapses into a straight line. A parabola needs the focus held off the line.

Two more names come straight out of the definition. The axis of the parabola is the line through $F$ perpendicular to $d$. Reflecting the plane across the axis fixes $F$ and maps $d$ onto itself, so it carries the whole curve onto itself: reflect any point that satisfies the distance condition, and the image still satisfies it. The vertex is the point of the curve on that axis. The vertex must be the midpoint between $F$ and the directrix, since that is the one point on the axis whose two distances agree.

To turn that picture into algebra you need to measure two distances, so here are the two measuring tools, both of which come from the Pythagorean theorem.

Distance between two points. Put $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ and complete the right triangle whose legs are horizontal and vertical. The horizontal leg has length $|x_2 - x_1|$, the vertical leg has length $|y_2 - y_1|$, and $PQ$ is the hypotenuse, so the Pythagorean theorem gives

Distance from a point to a line. The distance from a point to a line means the shortest distance, and it is measured along the perpendicular. Any other segment from $P$ to the line is the hypotenuse of a right triangle whose leg is that perpendicular. A hypotenuse is always longer than either leg, so no other route can be shorter. For a horizontal line $y = m$ the perpendicular is vertical, so the distance from $P = (x_0, y_0)$ is simply

and for a vertical line $x = m$ it is $|x_0 - m|$. Nothing more complicated is needed, because every directrix in this lesson is horizontal or vertical.

Deriving the equation from the definition

Now put the definition to work in coordinates. Place the vertex at the origin, and let $p$ be the signed distance from the vertex up to the focus, so that

The vertex $(0,0)$ is then correctly halfway between them. Everything about the curve is now encoded in the single number $p$.

Read the result again, because it is the punchline of the lesson. Dividing by $4p$ turns the equation into

which is $y = ax^2$ with $a = \dfrac{1}{4p}$. The purely geometric locus, defined without a single coefficient, is the quadratic graph you have been drawing since Chapter 4. Run the relation the other way and every quadratic $y = ax^2$ acquires a focus and a directrix it always had but never advertised:

For $y = x^2$, for instance, $a = 1$ gives $p = \tfrac14$: that familiar curve is the set of points equidistant from the point $\left(0, \tfrac14\right)$ and the line $y = -\tfrac14$.

The sign of $p$ carries the direction. If $p > 0$ the focus sits above the vertex, the directrix below, and the curve opens upward, wrapping around its focus. If $p < 0$ the focus sits below the vertex, the directrix above, and the curve opens downward. In both cases the parabola bends toward the focus and runs away from the directrix, which is the fastest way to get the direction right without memorizing anything.

The definition also hands you a shortcut worth keeping. For a point $(x,y)$ on $x^2 = 4py$ with $p > 0$, its distance to the focus equals its distance to the directrix by definition, and the second of those is easy:

No square root, no substitution. The distance from a point on the parabola to the focus, called its focal distance, is read straight off the $y$-coordinate.

The focal width, and why the number is $4p$

Where does the $4$ in $x^2 = 4py$ actually show up on the picture? Draw the chord through the focus parallel to the directrix and measure it. On $x^2 = 4py$ the focus is at height $y = p$, so substitute:

The chord runs from $(-2p,\, p)$ to $(2p,\, p)$, so its length is $|4p|$. That chord is the focal width, and it is the single most useful number for sketching a parabola by hand. Go to the focus, move $|2p|$ each way, and you have two more points on the curve.

This also settles the most common confusion in this lesson. The coefficient in $x^2 = 4py$ is $4p$, and it is not the focus distance. Its magnitude, $|4p|$, is the focal width; the focus distance is a quarter of that magnitude, $|p|$. A curve with a big $|4p|$ is wide and shallow, and its focus sits far from the vertex. A curve with a small $|4p|$ is narrow and steep, and its focus is tucked in close.

Moving the vertex

Nothing so far depended on the vertex being at the origin, and moving it costs no new work. Translation is a rigid motion: sliding two points by the same vector does not change the distance between them. In the same way, sliding a point and a line by the same vector does not change the distance from the point to the line. So if you translate the focus and the directrix by $(h, k)$, the locus of equidistant points translates by $(h,k)$ as well. That follows because the defining condition is stated entirely in terms of distances.

Apply that to the focus $(0,p)$ and directrix $y = -p$, whose locus is $x^2 = 4py$. Moving the vertex to $(h,k)$ means every distance is now measured from $(h,k)$ instead of from the origin, so $x$ is replaced by $x - h$ and $y$ is replaced by $y - k$ everywhere. The focus moves to $(h,\, k+p)$, the directrix becomes $y = k - p$, and the equation $x^2 = 4py$ becomes the standard form of a parabola with a vertical axis:

This is the bridge back to Chapter 4. Any quadratic $y = ax^2 + bx + c$ can be put into vertex form $y = a(x-h)^2 + k$ by completing the square, and rearranging that gives $(x-h)^2 = \tfrac{1}{a}(y - k)$, which is the standard form above with $4p = \tfrac{1}{a}$. So every quadratic you have ever graphed already carries a focus and a directrix inside its vertex form; completing the square is all it takes to read them off.

Notice the shape of that conversion: the square must stand alone on one side. As soon as you write $y = a(x-h)^2 + k$ you are one step away, but the step matters, because $4p = \tfrac1a$ and not $a$.

Parabolas that open sideways

Nothing in the definition says the directrix has to be horizontal. Take a vertical directrix instead, $x = -p$, with the focus at $(p, 0)$. The derivation is the same computation with the roles of $x$ and $y$ exchanged: a point $(x,y)$ is on the curve when

and squaring, expanding, and canceling the $x^2$ and $p^2$ terms leaves

Shifting the vertex to $(h,k)$ by the same translation argument gives the standard form of a parabola with a horizontal axis:

Now $p > 0$ opens the curve to the right and $p < 0$ opens it to the left, and the focal width is still $|4p|$.

$y^2 = 4px$ is not the graph of a function of $x$: solving for $y$ gives $y = \pm 2\sqrt{px}$, two values at every allowed $x$ past the vertex (just one, $y = 0$, exactly at the vertex). That is already enough for the vertical line test to fail, since it only takes one $x$ where a vertical line crosses the curve twice. It is still a perfectly good parabola under the geometric definition, which never mentioned functions. Turn your head sideways, though, and it is a function of $y$: $x = \tfrac{1}{4p}\,y^2$.

Why the focus deserves its name

Every ray that arrives parallel to the axis leaves the curve through one single point, which is exactly why that point is called the focus. A radio dish collects a plane wave from a satellite millions of kilometers away and concentrates it onto a receiver the size of a thumb. A telescope mirror gathers starlight and delivers it to the eyepiece. A car headlight runs the same idea backwards, putting the bulb at the focus so the reflected light leaves in a parallel beam.

Here is the plain reason it works. Reflection off a curve keeps equal angles to the tangent line at the point of impact. At any point $P$ on the parabola, $PF = PD$, so $P$ sits exactly halfway, in angle, between the focus $F$ and the foot of the perpendicular $D$ on the directrix: the tangent at $P$ is the line that treats $F$ and $D$ as mirror images of each other. A ray coming straight down the axis direction arrives along $PD$, bounces off that tangent, and leaves along $PF$, headed for the focus. Since this happens at every point of the curve, every ray parallel to the axis, wherever it lands, ends up at the same point $F$.

Common mistakes