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Parabolas: From Definition to Equation

Module by: Kenny M. Felder. E-mail the author

Summary: A teacher's guide on teaching the connection between the definition and equation of a parabola, and how to get from one to the other.

OK, where are we? We started with the geometric definition of a parabola. Then we jumped straight to the machinery, and we never attempted to connect the two. But that is what we’re going to do now.

Remember what we did with circles? We started with our geometric definition. We picked an arbitrary point on the circle, called it ( x x, y y), and wrote an equation that said “you, Mr. ( x x, y y), are exactly 5 units away from the origin.” That equation became the equation for the circle.

Now we’re going to do the same thing with a parabola. We’re going to write an equation that says “you, Mr. ( x x, y y), are the same distance from the focus that you are from the directrix.” In doing so, we will write the equation for a parabola, based on the geometric definition. And we will discover, along the way, that the distance from the focus to the vertex really is 14a14a size 12{ { {1} over {4a} } } {}.

That’s really all the setup this assignment needs. But they will need a lot of help doing it. Let them go at it, in groups, and walk around and give hints when necessary. The answers we are looking for are:

  1. x2+y2x2+y2 size 12{ sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}. As always, hint at this by pushing them toward the Pythagorean triangle.
  2. y + 4 y+4. The way I always hint at this is by saying “Try numbers. Suppose instead of ( x x, y y) this were (3,10). Now, how about (10,3)?” and so on, until they see that they are just adding 4 to the y-coordinate. Then remind them of the rule, from day one of this unit—to find distances, subtract. In this case, subtract -4.
  3. x2+y2= y + 4 x2+y2 size 12{ sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}=y+4. This is the key step! By asserting that d 1 = d 2 d1=d2, we are writing the definition equation for the parabola.
  4. Good algebra exercise! Square both sides, the y 2 y 2 terms cancel, and you’re left with x 2 = 8 y + 16 x 2 =8y+16. Solve for y y, and you end up with y = 1 8 x 2 - 2 y= 1 8 x 2 -2. Some of them will have difficulty seeing that this is the final form—point out that we can rewrite it as y = 1 8 ( x - 0 ) 2 - 2 y= 1 8 (x-0 ) 2 -2, if that helps. So the vertex is (0,-2) and the distance from focus to vertex is 2, just as they should be.

Warn them that this will be on the test!

Which brings me to…

Time for another test!

Our first test on conics. I go back and forth as to whether I should give them a bunch of free information at the top of the test—but it’s probably a good idea to give them a chart, sort of like the one on top of my sample.

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