# Parabola

Object type: Plane curve

### Definition

In $\mathbb{R}^2$, a *parabola* is a set of points $(x, y)$ satisfying
the equation $$y = ax^2 + bx + c$$ for some constants $a, b, c \in \mathbb{R}$ with
$a \neq 0$, that is, the parabola is the *graph* of the function $x \mapsto
ax^2 + bx + c$. Below the parabola $y = x^2$ is shown.

The lowest point of the parabola is called its *vertex*, and the vertical
line passing through this point is the *axis of symmetry* of the curve. Since
$$a x^2 + bx + c = a \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c,$$ the vertex
is $\left(-b/2a, c - b^2/4a\right)$ and the axis of symmetry is $x = -b/2a$. Below we
will only consider the parabola $y = a x^2$, since every other parabola (with axis of
symmetry parallel to the vertical axis) can be obtained from this by translation.

### Properties

Given a straight line $L \subset \mathbb{R}^2$ and a point $F \in \mathbb{R}^2$ with
$F \not\in L$, the set of points equidistant to $L$ and $F$ is a parabola. Indeed, consider
the line $y = -a$ and the point $(0, a)$. A point $(x,y) \in \mathbb{R}^2$ is equidistant
to $L$ and $F$ iff $\sqrt{x^2 + \left(y - a\right)^2} = y + a$ which is equivalent to $y =
x^2/4a$, which is a parabola. Conversely, every point on the parabola $y = a x^2$ is readily
seen to be equidistant to the line $y = -1/4a$ and the point $(0, 1/4a)$. The line is called
the *directrix* of the parabola, and the point is called the *focus*. Hence,
there is a one-to-one correspondance between pairs (line, point (not on the line))
and parabolae in the plane.

### Curvature

The curvature function for the parabola with vertex at the origin, that is, $y = a x^2$, parameterised using $x$ as parameter, is $$\kappa(t) = \frac{2 |a|}{\left(1 + 4 a^2 t^2\right)^{3/2}}.$$ Notice that the curvature is highest at the vertex.

### Parabolae as Mirrors

A light-ray travelling towards a parabolic mirror parallel to its axis of symmetry will be reflected in the direction towards the focus, independent of where the ray hits the mirror. This makes parabolic mirrors very useful in technical applications. Below we compare a parabolic mirror with a circular mirror. The circular mirror fails to concentrate the light-rays to a single point.

(You might also experience a Hering illusion in the first picture.)

**Proof of the focusing property of a parabolic mirror.** Consider the parabola
$y = a x^2$ and the particular light-ray $x = h > 0$ coming from above towards the parabola.
This light-ray will hit the parabola at $(h, ah^2)$. Given the scalar field $F(x,y) = y - ax^2$,
the parabola is the level curve $F(x, y) = 0$, and so its normal at $(h, ah^2)$ is the scalar
field's gradient at this point, namely, $\mathbf{n} = (-2ah, 1)$. Notice that the normal
points upwards. Let $\alpha$ be the angle between the vertical direction $\hat{\mathbf{y}}
= (0,1)$ and $\mathbf{n}$. Since $$\left\langle\mathbf{n}, \hat{\mathbf{y}}\right\rangle = 1 =
\sqrt{4a^2h^2 + 1} \cos{\alpha},$$ we have $$\alpha = \arccos{\frac{1}{\sqrt{4a^2h^2 +1}}}.$$
Notice that $\alpha$ is the angle between the normal of the reflecting line and the incoming
light-ray. By the law of reflection, this angle is the same as the angle between the normal
and the reflected line. Thus the angle between the vertical direction $\hat{\mathbf{y}}$ and
the reflected line is $x := 2\alpha$. Let $F_h$ denote the intersection of the reflected line
with the $y$-axis $x = 0$.

Consider now the reflected line $y = kx + m$. Let $\Delta x := h > 0$ and $\Delta y < 0$ be
the change in $x$ and $y$, respectively, on the reflected line from $x = 0$ to the point of
reflection at $x = h$. Since $\Delta x$ and $-\Delta y$ are the lengths of the catheti of the
right triangle with vertices at $F_h$, $F_h + h \mathbf{\hat{x}}$, and $(h, ah^2)$, we have
$$\frac{\Delta x}{-\Delta y} = \tan{x} = \tan{2\arccos{\frac{1}{\sqrt{4a^2 h^2 + 1}}}} =
\frac{4ah}{1 - 4a^2 h^2},$$ as is readily verified using elementary trigonometry. Thus,
$$k := \frac{\Delta y}{\Delta x} = \frac{4 a^2 h^2 -1}{4ah}.$$ Since $(h, ah^2)$ lies on the
reflected line, $$m = ah^2 - \frac{4a^2 h^2 - 1}{4ah} h.$$ Therefore, the reflected line is
$$y = \frac{4a^2 h^2 - 1}{4ah} x + ah^2 - \frac{4a^2 h^2 - 1}{4a}.$$ Putting $x=0$ yields
$y = 1/4a$; consequently, $$F_h = \left(0, \frac{1}{4a}\right),$$ which remarkably does
not depend upon $h$. Hence, *every* incoming ray of light (parallel to the axis of
symmetry) will pass through $(0, 1/4a)$ after reflection, and we recognise this point as the
focus of the parabola. ∎

### Other Examples

- A body (such as a ball) thrown at the surface of the Earth (or in any other constant field of gravity) will follow a parabolic trajectory if non-gravitational influence is neglected (such as air resistance).
- The so-called 'velocity profile' in fully-developed Poiseuille flow of a Newtonian fluid in a straight circular cylindrical tube is parabolic. More precisely, if the (dynamic) viscosity of the fluid is $\mu$ and the constant radius of the tube is $R$, then, in cylindrical coordinates $(r, \varphi, z)$ where the tube is $r = R$, the velocity of flow at $(r, \varphi, z)$ inside the tube is $\mathbf{v} = -\frac{k}{4\mu}(R^2-r^2)\mathbf{\hat{z}}$ where $k$ is the pressure gradient.