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- anonymous

Can somedbody help me with problem 3F-8, (b)? I can't understand the answer given...

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- anonymous

Can somedbody help me with problem 3F-8, (b)? I can't understand the answer given...

- jamiebookeater

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- NoelGreco

There are several somebodies out here who would be glad to help if you included a link to the problem itself.

- anonymous

Sure, sorry about that: the problem is in this pdf file

- Stacey

|dw:1359705070117:dw| Basically we end up with OB=y and AQ=x

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- Stacey

The point O is (0, 2y) and the point Q is (2x, 0),
giving the slope of the line to be -y/x.

- anonymous

Thank you very much for your answer. I still don't know if I got it, though. What I understood from the problem is that I should find a curve such that for whatever point P in the curve that I chose, if I took the tangent at that point, P would be at exactly the midpoint of the segment of that line that lies in the first quadrant. But looking at your curve, and the one from the solution given, it seems to me that if I move up (down), the upper part of the line would be shorter (longer) than the lower part.
Did I get the question wrong? Thank you!

- Stacey

|dw:1359951384469:dw| You seem to understand the problem correctly. The upper portion of my curve is definitely off, but I did manage to show a tangent for the lower section that should illustrate the concept.

- Stacey

|dw:1359952181692:dw| This graph might illustrate it better.

- anonymous

Thank you very much

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