Now that we’ve solved the problem using tangent lines and slopes, we can take a crack at it using a different method and counting the intersections between the circle and the parabola.

Consider the circle as it moves down the y-axis. When it is high enough, it has no intersections with the parabola. As it descends there comes a point at which it has exactly two intersections with the parabola. Then if it descends any further, it has four intersections with the parabola. The bounce point is the point at which it has only two intersections.

What we want to find is that middle case: with exactly two intersections. The circle and the parabola will intersect at points (x, y) that are on both curves, and thus satisfy both equations.

Be sure you understand what this means. For a fixed value of t (a fixed circle) this is a set of two equations in the two unknowns x and y, and the solutions are the intersections belonging to that value of t. For example, for the circle in the bottom diagram (which has t = 1.15), the equations would have four solutions.

Let’s try to solve these equations and see what happens. Note that our solutions will depend on t. That is, we are looking for expressions for x and y in terms of t.