Definition: A parabola is the set of points in the plane that are equidistant from a point (the focus) and a line (the directrix.)
The following exercise should help convince you that this definition yields the parabolas you are familiar with.
Exercise: Given a focus at (0,1) and a directrix y=-1, find the equation of the parabola. How to do it: draw a figure showing a generic point P on the parabola, with coordinates (x,y). Calculate its distance to the focus, its distance to the directrix, set those equal, and simplify. Or, for a more general result, do this exercise for a focus at (0,f) and a directrix y=-f.
Given the focus (F) and directrix (d), here is a method to construct any number of points on the parabola: choose a point T on d. Construct the perpendicular bisector of TF. Construct the perpendicular to d through T. The intersection of these two lines (P) is a point on the parabola. (Make sure you understand why.)
Exercise: With the help of dynamic geometry software, construct P as outlined above, then trace P as T moves, or create its locus, which is the parabola.
- Find other constructions of the parabola, given focus and directrix. (Here is one solution, inspired by the focus-directrix graph paper mentioned above: GeoGebra | Cabri. If you send me your Sketchpad construction, I'll post them here and credit you.)
- Construct a parabola given its axis of symmetry, its vertex, and one other point on it.
A light ray originating at the focus will be reflected on the parabola and continue in a direction parallel to the axis of symmetry. Likewise, a light ray coming in parallel to the axis of symmetry will be reflected to hit the focus.
That this works is readily proved using the above construction, if you assume a basic fact from optics: the angle of incidence equals the angle of reflection. The key to the proof is realizing that MP must be tangent to the parabola. Indeed, assume MP intersects the parabola at P and another point P'. Because P' is on MP, it is equidistant from F and T. Because it is on the parabola, it is equidistant from F and d, so if T' is the foot of a perpendicular from P' to d, we have P'T = P'F = P'T'. But the only way for P'T to equal P'T' is for T' to be T. Therefore P' is P, and MP is tangent.
Exercise: Prove the reflection property of the parabola, assuming that the angles of incidence and reflection are determined with respect to the tangent to the parabola at the point of incidence.
This property is of course the basis of many applications (headlights, flashlights, satellite dishes, radar...) For example, here is a diagram of how this works in a reflector telescope:
The primary mirror is parabolic, reflecting the parallel rays to the focus. The secondary (flat) mirror redirects this towards the eyepiece.
Like squares and circles, unlike rectangles and ellipses, all parabolas are similar. They cannot be "pointier" or "wider". They all have exactly the same shape, which appears "pointier" from afar, and "wider" when looked at in the neighborhood of the vertex.
Unfortunately, many of us have misled many students by implying otherwise: we often claim that changing the value of a in the formula y=ax2 changes the shape of the parabola. In fact, many teachers believe this to be true. Here are three types of arguments to show it is a misunderstanding.
- Algebraic Argument:
- In other words, in the equation y=x2, both x and y have been multiplied by the same number a. The parabola is scaled with no distortion.
Since the directrix is infinite, moving the focus has no effect on the parabola's shape. It is merely zooming in or out on one shape.
- Visual Argument:
- Same equation, apparently different shapes:
- (Dan Bennett suggests a dramatic illustration of this: make a transparency of a figure like the one above. Project it. Use another transparency to trace a piece of the projection, like the one below. Compare the two transparencies, which seem to have very different shapes, but clearly must represent the same equation. The same approach works on a document camera: place the original next to the projected image of a photocopy of itself.)
- In the illustration below, you can use the Zoom In or Zoom Out tool. Just click near the origin, and see how the same parabola's shape appears to change.
- Conversely, look at how different equations can yield the same shape:
- In fact, you can see for yourself: in the applet below, drag the axes' unit (the "1" on the x-axis) left or right, and watch the "a" in the equation change while the parabola's shape remains absolutely constant.
Finally, in this illustration, you can show that any two parabolas are similar by translating the first, and then dilating it, so that the final image is superposed on the second. (Use the sliders in order.)
A worksheet on Parabola Similarity. (Requires GeoGebra or other dynamic geometry software.)