Differential Geometry of Curves and Surfaces - Chapter 1 Section 3 Exercise 3

Problem:

Note the errata to this problem here. In particular, B should be on the line with OC and that's the half line r.

Solution:

Note that $ \angle OCA $ is a right angle.

$ \frac{CA}{2a} = \sin \theta $
$ \frac{BA}{2a} = \tan \theta $
$ CA^2 + CB^2 = BA^2 $

$ (2a \sin \theta)^2 + CB^2 = (2a \tan \theta)^2 $
$ OP^2 = CB^2 = 4a^2(\tan^2 \theta - \sin^2 \theta) $

With some trigonometry, we get

$ \sec^2 \theta = 1 + \tan^2 \theta = 1 + t^2 $
$ \cos^2 \theta = \frac{1}{\sec^2 \theta} = \frac{1}{1 + t^2} $
$ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{1}{1 + t^2} = \frac{t^2}{1 + t^2} $
$ \tan^2 \theta - \sin^2 \theta = t^2 - \frac{t^2}{1 + t^2} = \frac{t^4 + t^2 - t^2}{1 + t^2} = \frac{t^4}{1 + t^2} $

So we get $ OP = \frac{2at^2}{\sqrt{1 + t^2}} $

$ x = OP \cos \theta = \frac{2at^2}{1 + t^2} $
$ y = OP \sin \theta = \frac{2at^3}{1 + t^2} $

This time around I solved the problem with a good diagram and geometrical insight without using MATLAB.

Part (b) is trivial, when $ t \to \pm \infty $, $ x \to 2a $, $ y \to 0 $,

$ \alpha' = 2a\frac{1}{(1+t^2)^2}((1 + t^2)2t - t^2(2t), (1 + t^2)3t^2 - t^3(2t)) $

So $ \alpha'(t) = (0, 2a) $ as $ t \to \pm \infty $

At this point I am still confused, look like $ \alpha'(t) $ should be $ (0, 2a) $ and not $ (2a, 0) $?