It turns out, that if you pick three random points on a equilateral hyperbola (they can be on either branch), then the orthocenter will also fall on the hyperbola. Stated another way, if you pick the three points all on one branch and make them all free to move, the locus of the orthocenter will be the other branch of the hyperbola.
Poncelet had actually written about this as far back as Jan of 1821 in Gergonne's Annales.
Oh, by the way, a little "prove this factoid" for my calc kids... the y-intercept of the tangent line to any point on the rectangular hyperbola is always twice the y-coordinate, and the slope is always the square of the reciprocal of the y-coordinate. SWEET! [Yikes, I've been busted... Keninwa noticed a mistake in the above (thanks guy) actually what I should have said (and this is only true for the basic y=1/x case), the slope is equal to the negative of the square of the y=value .. (and now, head hanging in shame, he wanders off into the sunset, muttering to himself about proofreading)..
3 comments:
You found one thing you can do!
Your posts always make my head hurt...in a good way.
I was thinking of actually doing this with my calc kids, but these theorems are new to me so I wanted to clarify what I suspect might be a typo in your post. You state that the slope of the tangent is the recip. of the square of the y-coor. (a value that will always be positive). However, consider point A in your example; the x and y coor. are both positive, but a tangent line thru that point would obviously have a neg. slope (In fact, all the tangent lines for your hyp will have neg slopes). What determines the sign of the slope in your theorem?
Hi Pat,
I am in Midland. I like your blog; it is one of several that I read regularly.
Post a Comment