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arturop
 3 years ago
I don't understand the geometry of 3F  8b (not even what's been asked). If someone can help I'd be grateful.
arturop
 3 years ago
I don't understand the geometry of 3F  8b (not even what's been asked). If someone can help I'd be grateful.

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EulerGroupie
 3 years ago
Best ResponseYou've already chosen the best response.1A plane curve is just a curve in a plane... I would say that every function covered so far is a plane curve (2 dimensions... y and x). Bisecting means the half way point. So we are looking for some curve in quadrant 1 that can cut a line in half at its tangent point. I would say that the only line that can be cut in half won't shoot off to infinity in the first quadrant, so it must intersect both the postive yaxis and the positive xaxis. Here are a few possible lines.dw:1353866000981:dw Each line has a midpoint between the axes. The curve that touches all of those midpoints is the curve we are looking for. I hope that helps, because I have to go...

EulerGroupie
 3 years ago
Best ResponseYou've already chosen the best response.1One more thing. If we draw a ton of lines like that, we get the cool optical illusion where a bunch of lines make a curve... our curve. It looks like y=1/x to me (just guessing)... but I'm sure there is a way more analytical method.

arturop
 3 years ago
Best ResponseYou've already chosen the best response.0Got it now, thanks very much. You are right in that the solution is the family of hyperbolas. The solution is not very clear in my opinion either. I understand now but I am sure I am always going to struggle translating this kind of problems to equations...
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