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Geometry of the Parabola (2D)by Henri Picciotto | |
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Parabolas are a central topic in high school algebra classes, but, perhaps because of the rigid separation between algebra and geometry classes in the US secondary curriculum, we do not usually treat them as geometric objects. While most teachers are aware of some of the parabola's geometric properties, few of us are familiar with the proofs of those properties.
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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.
TopGiven 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. [The figure above was created in Cabri, where a conic section can be defined by five points. You can drag F or T. To replay the construction step by step, double-click it and use the toolbar that appears at the bottom.]
TopA 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. This property is of course the basis of many applications (headlights, telescopes, satellite dishes, ...)
This 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, if it intersected it again at a point P', that point would be equidistant from F and T, but it would necessarily be further from or closer to d, and thus could not be on the parabola -- a contradiction. So P' cannot exist, and MP is a 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.
TopLike 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.
Geometric Argument:
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.




AcknowledgmentsThanks to Kim Seashore for helping me think about "all parabolas are similar," and to Geneviève Tulloue for improving the Cabri applets. | |
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| Visit Henri Picciotto's Math Education Page. | Send me e-mail . |