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PhiNotPi's avatar

What is the equation of this curve? (see details)

Asked by PhiNotPi (11621 points ) October 7th, 2011

If you have two points (a,b) and (c,d), there are an infinite number of parabolas that can go through these points. If you plot all of the vertices of these parabolas, they create a curve. What is the formula for this curve?

Here is an image showing multiple parabolas through two points.

This is NOT homework.

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34 Answers

gasman's avatar

Hmm…interesting question. I’ll give you a quick answer based on my intuition but no actual analysis. I think any point in the plane can be chosen as the vertex of a parabola that passes through the other two given points. In other words, I don’t think the answer is restricted to a simple curve.

There might be points that are not in the set. For instance, a collinear point can serve as the vertex only if a straight line is considered a degenerate case of a parabola.

PhiNotPi's avatar

@gasman If there are two points (0,0) and (1,1), I am sure that any point (x>1, y<1) can’t be a vertex. Same thing with a point (x<0, y>0).

Mariah's avatar

[Grabs some paper]

I’ll be right back.

wundayatta's avatar

Would any point that is not in the plane of the line connecting the two points nor on the side of the tails of the parabola be a possible vertex of the parabola? There is no curve. It is a field or a space?

PhiNotPi's avatar

I am still pretty sure that there is a curve. If there is not a curve, then the resulting shape has to have some vertical thickness. This means that moving the vertex up or down (within a certain range) still makes it so that the resulting parabola still goes through both points.

CWOTUS's avatar

@wundayatta is onto something. There’s no particular equation or curve here. The solution is going to be a plane. Consider that the parabola doesn’t have to curve to a vertice – either high or low – between your points. (You’ve shown it that way, but the points may be on either leg of the parabola. There’s no telling where the vertice will be.)

PhiNotPi's avatar

Well, if there is a plane, what are the boundaries of the plane?

CWOTUS's avatar

There are no boundaries. The vertices could be anywhere on the plane.

PhiNotPi's avatar

If the two points are (0,0) and (1,1) what is the formula for the parabola with vertex (2,-1) that travels through those points? As far as I can tell, there isn’t one.

CWOTUS's avatar

Of course there is. Turn the parabola upside-down (from the way you’re envisioning it now) and imagine a vertice at (0.5, 2) or some other point above the three you’ve listed.

PhiNotPi's avatar

You said that the vertex could be anywhere on the plane, and I am using (2,-1) as an example. If this vertex can work, then there is a equation with this point as the vertex that goes through those two points.

Mariah's avatar

What I’ve been trying has been getting me nowhere, sorry. I’ll keep looking at it, though.

Mariah's avatar

I’ll throw out some tidbits that have been floating around my mind in case anybody could do something useful with them:

Parabolas can be written in two useful forms. The most relevant to what we’re doing is vertex form, which is in the form y = A(x-h)^2 + k where (h,k) is the vertex. In our case, we’d have to satisfy both b = A(a-h)^2 + k and d = A(c-h)^2 + k.

General form is y = Ax^2 + Bx + C, and similarly, both b = Aa^2 + Ba +C and d = Ac^2 + Bc + C must be satisfied. In this form, the vertex lies at (-B/(2A), C – B^2/(4A)).

I’ll keep thinking.

CWOTUS's avatar

What I meant – and what I presumed you understood – was that “given an infinite number of parabolas through the two given points”, vertices could and would be spread all over the plane.

Once you specify a third point, then we have to get a lot more particular about drawing the parabola to include “those three points”.

I’m going to disagree with @Mariah‘s assertion that the “x” value of the vertice has to be between the first two specified points. That’s the way @PhiNotPi has drawn his parabolas, but who says the vertice has to be between those points?

Mariah's avatar

Yeah I just deleted that 2 seconds after posting my response because I realized my error, haha. I was hoping nobody had seen it. ;)

I will amend my statement to say that the vertex cannot have the same y value as either of the points. The y value of the vertex is unique in all parabolas. So is the x value of the vertex, incidentally, so it also can’t have the same x value of either of the points. Unless, of course, one of the points is the vertex.

gasman's avatar

@Mariah The formulas you give are for a parabola whose axis (bisector) is oriented vertically. In general it might be oriented at some angle, in which case there will be xy terms as well.

Mariah's avatar

Hmm, I guess I assumed we were only talking about parabolas with vertical axes. That adds a large degree of difficulty if we have to consider others.

bobbinhood's avatar

Assuming we are talking about parabolic functions, it seems like we would be able to place a vertex anywhere except the closed horizontal strip between the two points, and except for the vertical lines through the two points. Although, the two points themselves could be vertices of various parabolas.

If we consider every parabola, regardless of whether it is a function, it seems like we could put a vertex everywhere except on the line containing the two points. Although, once again, the points themselves could be vertices.

At this point, I’m discussing this intuitively, not from any degree of proof.

Mariah's avatar

And if the points have the same value y, the vertex can only go on the vertical line that lies exactly between the points…right? Assuming again that we’re talking about parabolas with vertical axes.

bobbinhood's avatar

@Mariah I did not consider if the points had the same x- or y-values. That would certainly change things. My above answer only applies when neither the x’s nor the y’s are equal.

With what you’re saying (the y’s are equal and the axes of the parabolas are vertical), I would say that the vertex could go anywhere in the open vertical strip between the points, except for the horizontal line through the points. Even if we were to allow for parabolas with horizontal axes, I don’t think that changes our set of vertices.

Similarly, if the x’s are equal, we end up with the open horizontal strip between the points minus the vertical line between the points, but only if we consider all parabolas. If we only consider functional parabolas (those with vertical axes), then we end up with the empty set.

…I think.

bobbinhood's avatar

I edited the above answer because I typed too fast and some of what I said didn’t make sense. Just letting you know in case you need to refresh to see the actual answer before you respond.

bobbinhood's avatar

My first answer should have said “open horizontal strip” not “closed horizontal strip,” but it’s too late to fix it.

gasman's avatar

I thought we were talking about general functions F(x,y) such that the graph of F(x,y) = 0 is a parabola. This won’t necessarily be a function y = f(x) that is single-valued for every x.

@PhiNotPi If there are two points (0,0) and (1,1), I am sure that any point (x>1, y<1) can’t be a vertex. Same thing with a point (x<0, y>0).

I wish I could just post a diagram. I can sketch lots of parabolas that contradict the above conditions. For starters, any point on the line y = -x + 1, which is the perpendicular bisector of these points, can serve as a vertex of a parabola through those points. More generally, though, such symmetry isn’t required by the stated problem.

Mariah's avatar

@bobbinhood “With what you’re saying (the y’s are equal and the axes of the parabolas are vertical), I would say that the vertex could go anywhere in the open vertical strip between the points, except for the horizontal line through the points.”

I really think it could only go on the vertical line exactly in between the two points. Correct me if I’m wrong, but I believe all parabolas with vertical axes have vertical symmetry – so the x value of the vertex would have to be exactly in the middle of the two points.

“If we only consider functional parabolas (those with vertical axes), then we end up with the empty set.” Agreed, as it’s not a function if any two points have the same x value.

Mariah's avatar

@gasman Functions, by definition, are single-valued for every x. Or am I misunderstanding you?

bobbinhood's avatar

@Mariah You win. You would think I would remember that since I just proved it to my class last week.

Mariah's avatar

Another limitation: no vertices can exist that have y values between the y values of the two points, as the vertex of a parabola is always an absolute maximum or an absolute minimum.

CWOTUS's avatar

@bobbinhood

If either of the points can be a vertex of one or many parabolas, then by definition “the line that contains those two points” is not excluded as potential loci for vertices.

Mariah's avatar

@CWOTUS The lines are excluded except in the case that a vertex is on one of the points. No points on a parabola share the x or y value of the vertex.

CWOTUS's avatar

If the points are in a horizontal or vertical line then I’d agree with you, @Mariah, that the line between the two points is excluded. Your statement is correct, that “no points on a parabola share the x or y value of the vertex.” (In point of fact, if the parabola is a function, then no point can share the x axis value anywhere on the parabola. And the y value of the vertex is unique as the top or bottom of the curve.)

gasman's avatar

@Mariah Yes, a function f(x) is single-valued so the graph y = f(x) has certain restrictions that exclude, for instance, horizontal parabolas. But @PhiNotPi is asking about all parabolas that pass through two points. This need not be restricted to the form y = f(x).

To give a different example, the graph of x^2 + y^2 – 1 = 0 is a circle. It is not a single-valued function y = f(x). But you can consider G(x,y) to be a function of two variables where G = x^2 + y^2 – 1 whose graph G=0 is a unit circle centered at the origin.

I don’t remember the general form of a parabola and if I had time I could work out a transformation of axes allowing a vertical parabola (such as you described earlier) to be transformed to a tilted parabola whose axis is along any arbitrary line in the plane, in which case G(x,y) will contain xy terms—this is true of any conic section..

Speaking of planes, however, I am on the eve of international travel & might not have time to work out a more complete answer! (Hate when that happens.) I know that’s a cheap cop-out, though it happens to be true. Maybe somebody else could work it out.

Mariah's avatar

@CWOTUS Oh, I see what you’re saying now. Yeah, no excluding a diagonal line between the two points. But the vertical and horizontal lines that contain each of the points is off limits.

@gasman Yeah, I just went into this assuming we were talking about vertical parabolas. Maybe that’s not what @PhiNotPi wanted; he’ll have to answer that.

LostInParadise's avatar

It takes four points to determine a parabola. I suspect that any point can be chosen as a vertex. The problem becomes interesting if you only consider parabolas with the axis parallel to the y axis. In that case, 3 points determine a parabola and obviously only some of these will be vertices.

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