Consider: the circle of radius 2, center (0,2) The point P=(x, y) is found as fo

Question

Consider: the circle of radius 2, center (0,2) The point P=(x, y) is found as follows: Choose a point A on the line y=4 and connect it to the origin with a line segment which intersects the circle in a point B. On the line segment OB find the point R whose distance to the origin is 2/3 of the distance of B to 0. P is the point where the vertical line through B intersects the horizontal line through R. The point P traces a curve as A moves along the line y=4. Find a parameterization of this curve by expressing the coordinates x=x(theta) and y=y(theta) of p in terms of the angle theta, the radian measure of the angle that the line segment OA makes with the positive x-axis.

Explanation / Answer

If the angle is theta, B has coordinate (z cos theta, z sin theta). As B is on the circle of radius 2 and center (0,2) (z cos theta - 0)^2 + (z sin theta - 2) ^ 2 = 4 Multiplying through, z^2 cos^2 theta + z^2 sin^2 theta - 4 z sin theta + 4 = 4 As cos^2 theta + sin^2 theta = 1, z^2 - 4 z sin theta = 0 z^2 = 4z sin theta Dividing through by z, z = 4 sin theta, so the (x, y) coordinate is (4 sin theta cos theta, 4 sin^2 theta) Since R is 2/3 of the way from the origin to B, R = 2/3(4 sin theta cos theta, 4 sin^2 theta) = (8/3 sin theta cos theta, 8/3 sin^2 theta) Then, P has the y coordinate of R and the x co-ordinate of B, so P = (4 sin theta cos theta, 8/3 sin^2 theta)

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