抛物面反射

Given a point, called the focus, and a line not containing the point, called the directrix, a parabola is defined to be the set of all points in the plane determined by the focus and the directrix that are equidistant from the focus and directrix.

Most of us might be familiar with a different definition of a parabola, perhaps a graph of a quadratic function. Let us see how these $2$ definitions coincide.

Consider the parabola $y=x^2$. Try to find the focus and directrix. As a hint, the focus clearly must lie on the parabola's axis of symmetry, and the directrix must be perpendicular to the axis of symmetry.

Consider a point $P=(p,p^2)$ on the parabola. Let $F=(0,\frac{1}{4})$ and the line $l$ be the line $y=-1/4$. Then, the distance from $P$ to $l$ is $p^2+\frac{1}{4}$, while the distance $PF$ is $\sqrt{p^2+\left(p^2-\frac{1}{4}\right)^2}=p^2+\frac{1}{4}$. Indeed, we see that $\left(0,\frac{1}{4}\right)$ and $y=-\frac{1}{4}$ are our desired focus and directrix, respectively.

In general, if you are given a focus at $(0,p)$ and a directrix at $y=-p$, then one can easily use the distance formula to derive the equation of the parabola: $$\sqrt{x^2+(y-p)^2}=y+p$$ $$x^2+(y-p)^2=(y+p)^2$$ $$y=\frac{1}{4p}x^2$$

What do parabolas have to do with reflection? Suppose that you lie "above" the parabola and are looking down at the "bowl" of the parabola. You shoot a beam of light directly down so that it reflects off the surface of the parabola. It turns out that, no matter where you are, the beam of light will reflect off the parabolic surface so that it hits the focus.

To prove that this is true, let $P=\left(a, \frac{1}{4p}a^2\right)$ be an arbitrary point on the parabola, let $F=(0,p)$ be the focus, let $d$ be the line of the directrix: $y=-p$, and let $D=(a,-p)$ be the projection of $P$ onto the directrix.

The slope of the line $PF$ is $\frac{\frac{1}{4p}a^2-p}{a-0}=\frac{a^2-4p^2}{4ap}$. Thus, if $\theta_1$ is the angle between $PF$ and the positive $x$-axis, then $\cot\theta_1=\frac{4ap}{a^2-4p^2}$.

The slope of the line $t$ tangent to the parabola at $P$ is $\frac{a}{2p}$. Thus, if $\theta_2$ is the angle between $t$ and the positive $x$-axis, then $\cot\theta_2=\frac{2p}{a}$.

Finally, the line $PD$ is parallel to the $y$-axis. Thus, if $\theta_3$ is the angle between $PD$ and the positive $x$-axis, then $\cot\theta_3=0$.

Let $\alpha$ be the acute angle between $t$ and $PD$, i.e. the angle of incidence, and let $\beta$ be the acute angle between $t$ and $PF$, i.e. the desired angle of reflection. We want to prove that $\alpha=\beta$.

We will use the following angle-subtraction formula: $$\cot(A-B)=\frac{\cot A\cot B +1}{\cot B -\cot A}$$ from which it follows that: $$\cot\alpha=\frac{0\frac{2p}{a}+1}{\frac{2p}{a}-0}$$ $$\cot\alpha=\frac{a}{2p}$$ $$\cot\beta=\frac{\frac{2p}{a}\frac{4ap}{a^2-4p^2}+1}{\frac{4ap}{a^2-4p^2}-\frac{2p}{a}}$$ $$\cot\beta=\frac{\frac{a^2+4p^2}{a^2-4p^2}}{\frac{2a^2p+8p^3}{a(a^2-4p^2)}}$$ $$\cot\beta=\frac{a}{2p}$$ It follows that $\alpha=\beta$. Therefore, if $\alpha$ is the angle of incidence, then $\beta$ will be the angle of reflection. It follows that a beam of light travelling parallel to the axis of symmetry of a parabola will be reflected to the focus by the parabola.

Now, suppose that we rotate the parabola around its axis of symmetry. The resulting surface is called a circular paraboloid. By the circular symmetry of this figure, we can easily see that this surface will have exactly the same property described above.

This fact is applied in many places. For example, many solar cookers are made in the shape of a circular paraboloid to optimize the use of the Sun's rays by focusing them all in one place. In addition, radio receivers are also in the shape of a circular paraboloid, since most signals that hit the dish will be from a constant direction. Thus, when the radio receiver is facing in the correct direction, any radio wave that hits the dish will be sent to the focus, where it can be collected and interpreted.

A paraboloid has the special property that any light ray from infinity has the same path length if it passes through the focus of the paraboloid.