Question

Problem Set 5 3F-8b. Point Can Bisect a Curve?

Answer 1

Problem Set 5, Problem 3F-8b reads:
b) Find all plane curves in the first quadrant such that for every point P on the curve, P bisects the part of the tangent line at P that lies in the first quadrant.

I was wondering if anyone could better explain what the question is asking. It doesn't seem like a point (P) can "bisect" a line. Don't lines bisect lines? To me bisecting implies a cut… I suppose a point can bisect a line if it lies on the line. If that's the case though, then doesn't every Point <x0, f(x0)> "bisect" the tangent line y-f(x0) = f'(x0)(x-x0)??

Or… is this maybe saying to find all the curves f(x), such that the line starting at the origin with slope f'(x) "bisect".

Thanks!

Answer 2

We're looking at lines that are tangent to the curve, and specifically the portion of the line that lies in the first quadrant, and we want the point of tangency to be the midpoint of that line segment. Consider a circle with radius sqrt(8) centered at the origin, for example. The point (2, 2) lies on the circle, and if you draw a tangent at that point, you'll see that (2, 2) is the exact midpoint of the segment of that tangent that lies in the first quadrant. If you move to a different point on the circle, however, the point of tangency won't be at the midpoint of the line segment created from a tangent at that point. The problem is asking you to figure out what kind of curve would have the property that every point on the curve has the property that it bisects (equally divides) the line segment composed of the tangent at that point within the first quadrant.

Answer 3

Ahhh, in other words, the point P should be the midpoint of the segment created by taking the x-intercept and y-intercept of the tangent line?

Answer 4

That's it!

Answer 5

Thanks so much again Creeksider!