What do you make of this time-eater for a bonus on a precalc final?

Find the locus of points, the sum of whose distances from a given point and a given line are constant.

• I think my wording is awkward. Better ideas?
• I am not sure what the figure is, though I am fairly confident it does not have corners.
• I was looking for any progress: a loose sketch, some application of the distance formula, almost anything graphic, analytic or algebraic.

So what do you think?

19 Comments leave one →
June 15, 2007 am30 4:15 am 4:15 am

You can do it in Cartesian coordinates. Let your point be the origin and the line be given by y=a. Then the set you want is given by

x^2 + y^2 + (y-a)^2 = b.

Completing the square gives

x^2 + 2(y – a/2)^2 = b – (a^2)/2

So depending on the relationship between your constant and the distance from the point to the line, this is either an ellipse, a point, or hyperbola.

June 15, 2007 am30 4:17 am 4:17 am

I guess the last possibility is the empty set and not a hyperbola.

3. June 15, 2007 am30 5:52 am 5:52 am

This doesn’t seem right. Can we really describe an ellipse with one line and one point?

Correct me if I am mistaken, but shouldn’t the first equation be: $x^2 + y^2 + |y - a| = b$ ?

June 15, 2007 am30 6:08 am 6:08 am

I think that’s a great problem. The wording is pretty awkward, though. Maybe something like “Given a line L, a point P, and a distance d, find the locus of points for which the sum of the distances to L and to P is always equal to d.”? That’s just off the top of my head, it definitely needs some careful editing.

I’ll use the same setup as Math grad, with the line y=a and the origin as the distinguished point. An equation describing the desired criteria is $|a-y| + \sqrt{x^2 + y^2} = b$.

This is where Math grad went wrong: you can’t just square both terms on the left independently. Moving |a-y| to the right side and then squaring, we get $x^2 + y^2 = \left (b - |a-y| \right)^2$

Expanding and doing a bunch of cancelling and collecting, etc. we get: $x^2 + (2a - 2b) y = (a-b)^2$, when $latex y a$.

These are both parabolas, with vertices at $(0,(a-b)/2)$ and $(0,(a+b)/2)$ respectively. They “collide” when y=a, at the points $(\pm \sqrt{b^2 - a^2}, a)$ — so there are sharp corners!

I wish I could draw a graph to illustrate, but hopefully you can get the idea from the above analysis.

5. June 15, 2007 am30 9:21 am 9:21 am

You are on wordpress. Make a dummy post, and use it to preview your LaTeX. (That’s what I do). or tell me in words what that should be (it came up bad again), and I will fix it.

June 15, 2007 pm30 4:06 pm 4:06 pm

ohhhh, I get it. It’s the less than and greater than signs screwing with the HTML. Sorry! OK, for real this time!

x^2 + (2a – 2b) y = (a-b)^2, when y < a, and

x^2 + (2a + 2b) y = (a+b)^2, when y > a.

Feel free to delete my munged posts.

7. June 15, 2007 pm30 4:27 pm 4:27 pm

Several “munged posts” deleted! Btw, I rescued the first two (there were four) from the spam filter. It really is a glitchy thing.

June 15, 2007 pm30 6:04 pm 6:04 pm

oh, I was wondering why those took a while to show up. I just figured it was moderated. Strange, I’ve never had any problems with the spam filter on my blog… well, not that I know of at least. =)

9. June 16, 2007 pm30 3:37 pm 3:37 pm

Jonathn–
Thank you for the kind words and recommendation…
As usual, you have devised a brilliant variation on a standard locus problem. The fact that you presented this without having completely solved the problem yourself creates an environment in which the teacher and the students together form a ‘research team’. What an amazing environment in your class! I suspected it might be a parabola because one of the quadratic terms would probably ‘cancel out’ just as in the standard focus-directrix definition. When I saw Brent’s solution, a few thoughts were triggered in my brain:
1. Would there even be a locus if b were less than a?
2. If b = a, I visualized a segment but I wasn’t sure.
3. Most importantly, if it was indeed a parabola, what would the focus be? (I suspected it would be the fixed point). The directrix?

Rather than use the origin as the fixed point, I decided to use the x-axis as the fixed line and (0,a) as the fixed point. I felt this would make the absolute value expression easier to work with. I didn’t immediately realize that this would SOLVE the directrix problem!
Ok, here goes the analysis:

(Assume b is greater than ‘a’ and ‘a’ is greater than 0)
(Since I’m still learning LaTeX and I don’t feel like looking up the html code or trying to preview this first, pls excuse the klutzy notation):

SQRT(x^2 + (y-a)^2) + |y| = b
SQRT(x^2 + (y-a)^2) = b – |y|
Assuming that b is greater than y and y is greater than 0, this equation defines part (an arc) of a parabola with focus (0,a) and directrix y = b!
If y is less than 0, b – |y| = b + y = y – (-b), assuming y is greater than -b. This defines part of a second parabola with the same focus (0,a) but directrix y = -b!

The other cases are interesting too. If b is less than a, then I believe the locus is empty. I put the proof in the margin of my notes and i seem to have misplaced it…

Seriously, pls correct all of my usual careless errors, although the result makes sense to me. I would ask students what the significance of your ‘fixed line’ is, since it’s not the directrix. What exactly is it in general? I believe Brent’s equations would help here.

Anyway, thanks for the brilliant problem (ok, puzzle), jonathan. Students will greatly benefit from this kind of challenge, particularly learning how to be creative in their choice of an origin and axes. Locus problems are beautiful and I know we have deemphasized them in our curriculum. Bring ’em back, pls! (How about making it part of a standardized national curriculum!).

10. June 16, 2007 pm30 8:14 pm 8:14 pm

I like making things up. But this is math, locus, distances… I am convinced someone asked this question before. And maybe there is even a name for these two intersecting parabolas, scaled at (a-b)/(a+b) in relation to each other?

Investigating for fun, is, mm, fun.

However, Dave, I didn’t work on this with my students. I simply offered it as a time wasting bonus on their final exam.

June 17, 2007 am30 4:01 am 4:01 am

I don’t think of it as a “time wasting” problem! How many of your students attempted the problem? What were their thoughts/methods?

12. July 3, 2007 pm31 5:38 pm 5:38 pm

it’s odd. The fault might be all of those books by Martin Gardner – I am sure I never learned it at school – but I just knew it was a parabola. Recreational math pays a lot :-)

13. December 1, 2007 pm31 11:37 pm 11:37 pm

D750Fg Test myfunction comment

14. June 24, 2008 pm30 5:26 pm 5:26 pm

RECREATIONAL math??!!!!

That’s almost as funny as recreational social studies.

15. June 25, 2008 am30 8:26 am 8:26 am

It’s what we call puzzles. I get 100,000 google hits. I am guessing that Martin Gardner is considered the king of recreational math…

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