Thursday, February 10, 2011

I'm Holding Somebody Responsible For This!

Step 1: Choose values for 'a' and 'c'.
Step 2: Move the 'b' slider from -10 to 10.
Step 3: Select all the points step 2 puts in the spreadsheet. Right click and select "Create list of points."
Step 4: Click on the "Conic through 5 points" tool.
Step 5: Click on 5 of your points generated in step 3.
Step 6: Right click on the resulting function in the algebra window and change standard form
Step 7: Compare this function to the Starting Function
Step 8: Explain to me why I never knew this before.

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keninwa said...


After a couple minutes of fiddling, I believe it does work for all real c, but a must be 1.

David said...

Are you sure? I just checked it with a=2.5.

Sue VanHattum said...

I selected the points. Right-click for me (mac) is control-click. Where is my cursor when I control-click? I couldn't get what I was supposed to.

David Cox said...

If you've selected the points and control-click do you get a menu that has something like "create list of points" pop up? Basically you want to take those points that show up in the spreadsheet and graph them, then do a quadratic regression. I'm not familiar with Mac, so I'm afraid I can't be more help.

Let me know if that works, though.

Sue VanHattum said...

I had tried to control-click in a few different places, before I wrote my first comment, and nothing happened. It might be because this computer is so old. I'll try it on my desktop computer.

Sue VanHattum said...

I finally tried it on my (newer) desktop mac. I had to select one point (lower left), go to upper right and hover over that point, and then use control (equivalent of right, with click) AND shift (to select from previous selection to current). I finally got it.

Back to the math. Let me see if I have it: If you fix a and c, and vary b, taking the vertices of the family of parabolas given by varying bs gives you a parabola with the came c and opposite a?

(Let's call that a preliminary hypothesis, while I go play with this more.)

Sue VanHattum said...

Why does the conic generator need 5 points? Don't we need just 3 for a parabola? I'm guessing an ellipse or hyperbola needs 4, since there are 4 parameters (is that the best word?) in (x-h)^2/a^2 +-(y-k)^2/b^2=1. Maybe the 5th helps pick which conic makes sense.

David Cox said...

Yup, you got it, Sue. I'm not sure why the conic tool requires five points, though. I know three points are enough to derive a quadratic equation, but doesn't that assume it's a quadratic? What about circles, ellipses or hyperbolas? Maybe 5 points are necessary to rule out the other options?