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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.

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  1. EulerGroupie
    • 3 years ago
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    A 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 y-axis and the positive x-axis. Here are a few possible lines.|dw:1353866000981:dw| Each line has a mid-point 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...

  2. EulerGroupie
    • 3 years ago
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    One 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.

  3. arturop
    • 3 years ago
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    Got 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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