Parabolas

The parabola as a locus

The distance from a point to a line is always measured along the perpendicular. In the figure, $F = (0, 2)$ and the directrix is $y = -2$. The matching dashed segments have equal length: 4 and 4 for the point on the right, 6.5 and 6.5 for the point on the left.

The vertex $V$ is the point of the parabola closest to the directrix. It sits exactly halfway between $F$ and the directrix. The line through $F$ perpendicular to the directrix is the axis of symmetry.

Check with the point $P(4, 2)$ from the figure: $4^2 = 16$ and $8 \cdot 2 = 16$. ✓

What if the directrix is slanted? Then you need the distance from a point to a line.

The $xy$-term signals that the axis is tilted. A tell-tale sign of a parabola is that the second-degree terms form a perfect square: here $x^2 + 2xy + y^2 = (x + y)^2$.

Standard form with the vertex at the origin

Place the vertex at $(0, 0)$ and the focus at $(p, 0)$, so the directrix is $x = -p$. The definition $PF = $ distance to directrix gives

The $x^2$ and $p^2$ terms cancel, which is why the equation is so clean.

Translated parabolas and general form

If the vertex is at $(h, k)$ instead of the origin, shift the whole picture: replace $x$ by $x - h$ and $y$ by $y - k$.

Drag the sliders. Here $a = \dfrac{1}{4p}$, so a small $a$ means a wide parabola with a large focal length.

Multiply out a translated parabola and you get a general form. For example, $(y - 2)^2 = 6(x - 1)$ becomes $y^2 - 4y - 6x + 10 = 0$. One variable is squared, the other appears only to the first power, and there is no $xy$-term. To go back, complete the square in the squared variable.

The reflective property

This is why satellite dishes put the receiver at the focus, why headlights put the bulb there, and why reflecting telescopes use parabolic mirrors. A satellite is so far away that its signal arrives as parallel rays. At the end of this page you will see why the property is true.

Modeling with parabolas

Arches, projectile paths and cables are often modeled by parabolas. The key move is to choose coordinates smartly and then find the equation from the given information.

A parabola with a vertical axis is also an Algebra 1 quadratic function. From $(x - h)^2 = 4p(y - k)$ we get

which is vertex form with $a = \dfrac{1}{4p}$. So if you know the vertex $(h, k)$ and one more point, use $y = a(x - h)^2 + k$ and solve for $a$. If you know three points, use $y = ax^2 + bx + c$ and solve a system.

Lines and parabolas: intersections

To find where a line meets a parabola, substitute the line's equation into the parabola's. You get a quadratic whose discriminant tells you what happens.

• $\Delta > 0$: the line crosses the parabola at two points (a secant line).

• $\Delta = 0$: one point, and the line is tangent (see the next section).

• $\Delta < 0$: the line misses the parabola.

If the two intersection points are $A$ and $B$, the segment $\overline{AB}$ is a chord. Its length comes from the distance formula.

Tangent lines

A tangent line touches the parabola at exactly one point without crossing into it. Algebraically: substitute, and require $\Delta = 0$ (a double root). Slide $k$ below until the line $y = x + k$ just touches the parabola $y = \tfrac{x^2}{4}$.

Check it: $\tfrac{x^2}{4} = x + k$ gives $x^2 - 4x - 4k = 0$, and $\Delta = 16 + 16k = 0$ gives $k = -1$.

When the point of tangency is known, there is a direct formula.

Tangency in optimization and design

Tangents solve two kinds of "find the best" problems.

Closest point to a line. Draw lines parallel to the given line and slide them toward the parabola. The first one that touches it is tangent. So the closest point is where the tangent is parallel to the given line.

Tangent to a given line. If a parabola has an unknown parameter and must be tangent to a line, substitute and set $\Delta = 0$.

Why the reflective property works

Take $y^2 = 4px$ and a point $P(x_0, y_0)$ on it. Let $T$ be where the tangent at $P$ meets the axis.

Reflection. An incoming ray parallel to the axis is parallel to $\overline{TF}$. By corresponding angles, the angle it makes with the tangent equals $\angle PTF$. Because $TF = PF$, the base angles of the isosceles triangle are equal: $\angle PTF = \angle TPF$. So the angle between the incoming ray and the tangent equals the angle between the tangent and $\overline{PF}$. That is exactly the law of reflection (angle in = angle out), so the reflected ray travels along $\overline{PF}$, straight to the focus.

Projections of the focus. Since $FT = FP$, the point $F$ lies on the perpendicular bisector of $\overline{TP}$. So the foot $M$ of the perpendicular from $F$ to the tangent is the midpoint of $\overline{TP}$:

Its $x$-coordinate is always 0, so every such foot lies on the $y$-axis, the tangent at the vertex. Some problems write the parabola as $y^2 = 2px$, where $p$ is the distance from the focus to the directrix. That $p$ is twice ours, but the argument is identical.

Back to the start. A satellite is so far away that its signal arrives as parallel rays. By the reflective property, every ray bounces to the focus. That is where the receiver goes, and it is why the dish must be a parabola and not just any curve.

Summary