Parameterizing a curve in R^2

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A circle with radius a centered at the origin O. T is a point on the circle such that OT makes angle t with the positive x-axis. The tangent to the circle at T meets the x-axis at X. The point P=(x,y) is at the intersection of the vertical line through X and horizontal line through T.

I need to parametrize the curve C traced out by P with angle t as the independent variable. (Let O denote the origin).

My approach was this:

The x component of P is the same as the x component of X, and the y component of P is the same as the y component of T.

so y = a*sin(t), which is correct according to the answer key.

To find the vector OX, I decided to find vectors OT and TX, then OT + TX = OX.

OT = [a*cos(t) , a*sin(t)] is easy to find.

TX would be perpendicular to OT and have a length from T to X.

TX = [a*sin(t) , -a*cos(t)] does not work since it has a fixed length (only works for multiples of t= pi/4)

TX = [sec(t)/a , -csc(t)/a] is the only other vector I could find that satisfies OT.TX = 0,

but OT + TX = OX = [a*cos(t) + sec(t)/a , a*sin(t) - csc(t)/a] has a y-component that is generally not 0.

My problem is I don't know how to find TX, which means I can't find OX, and hence the x-component of P in regard to t.

The answer key says P = [a*sec(t) , a*sin(t)]

2 Attachment(s)

Re: Parameterizing a curve in R^2

Hi,

Just draw a picture and it should be obvious from elementary trigonometry. The first attachment shows this for T in the first quadrant. There are similar diagrams for other locations of T. The 2nd draws the curve in question.

Attachment 29914

Attachment 29915