I was tutoring a student this afternoon in Algebra II who was learning about parabolas. One exercise asked her how she can tell from a parabola's equation if the parabola opens up, down, left or right. She asked me if parabolas could open diagonally. I told her it was an excellent question and that I was not sure. Specifically there were two things I was not sure about:
1. Does the definition of a parabola allow for a diagonally opening parabola?
2. Given a focus and a diagonal directrix, would it be possible to derive an equation for that parabola? What would the equation look like?
As to the first question, the definition that popped into my head during the tutoring session was that a parabola is a conic section (a curve generated by "slicing" through a conic: an infinite double cone). The Greeks who studied mathematical constructs as pure geometry certainly thought of parabolas that way. Our diagonally opening parabola would be just fine with this definition. We wanted to know if we could just rotate it about in the Cartesian plane. Having never studied or taught parabolas in any but horizontal or vertical orientations, I thought perhaps there was a precise definition that for some reason disallowed other orientations. The student and I agreed to both try looking it up after our session.
Regardless of whether or not it could be defined as a parabola, I was curious as to what its equation would look like, so my student and I started playing around. We drew a particularly simple diagonally opening parabola: One with a directrix of y = -x and a focus of (2,2).
To derive an equation, we could begin by saying that any point P(x,y) must be equidistant from the directrix and the focus. So PF = PN (where N is the closest point on the directrix)
Using the distance formula, we can easily write out PF:
PF = sqrt((x - 2)^2 + (y - 2)^2)
PN is a little trickier and during the tutoring session I couldn't think how to write an expression for it involving only x and y. But, as almost always happens to me if I am stumped by something while tutoring, the solution occurred to me as I drove home. I solve a lot of math problems while driving; I often even create visual proofs which my husband worries must mean I am not devoting as much attention as I should to my driving, but I think it is an entirely separate part of my mind and does not interfere with my driving.
Anyway, here's what I figured out while driving home:
Let N = (x_2, y_2). Since N is on the directrix whose equation is y = -x, we can say that N = (x_2, -x_2).
The shortest distance between P(x, y) and N(x_2, -x_2) lies along a line with slope 1. (This line is perpendicular to the directrix and so must have a slope which is the negative reciprocal of -1, i.e. 1)
So we can write an equation in point-slope form:
(y + x_2) = 1(x - x_2)
So we can solve for x_2 in terms of x and y! x_2 = (x - y)/2
So, we can now apply the distance formula to P(x, y) and N((x-y)/2, (y-x)/2)
sqrt((x - ((x+y)/2))^2 + (y - ((y-x)2))^2)
Setting PF equal to PN, we get:
sqrt((x - 2)^2 + (y - 2)^2) = sqrt((x - ((x+y)/2))^2 + (y - ((y-x)2))^2)
After a few rounds of simplifying, we get:
(-1/2)x^2 + xy + (-1/2)y^2 + 4x + 4y - 8 = 0
A lot more complicated-looking than a nice y = a(x-h)^2 +k! And this was a very simple diagonally opening parabola!
Anyway, I then looked into the formal algebraic definition of a parabola and was delighted to find that the diagonally opening parabola does satisfy it! Here it is:
A parabola is an equation of the form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 where B^2 = 4AC and A and C are not both zero.
In the case of our example B^2 =1 and 4AC = 4(-1/2)(-1/2), so yes! B^2 = 4AC! And neither A nor C is zero so we meet that condition as well.
Now that I know that a parabola can open in any direction, it seems obvious and I almost feel silly that I didn't know that before (since I am tutoring math for a living!), but no, I don't feel silly, I feel delighted, awed, and proud that mathematics is such a rich and fascinating subject and that I can continue to discover new things every day, hopefully also inspiring my students and giving them a taste for the joy of discovery.
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