If I understand the relationship (no guarantees) then there is a unique solution to each point.

Problem: A circle centered at the origin with a radius of r shares common tangents with a parabola y

^{2}= ax. If the tangent contains the point (p,q) on the circle; find the coordinates of the tangent point (s,t) on the parabola?

answer below

a

b

c

d

e

f

g

h

i

j

k

l

1

2

3

Ok.. here is the things I noticed...

a) For the common tangent to a circle x

^{2}+y

^{2}=r

^{2}at (p,q) and a parabola, y

^{2}= ax at (s,t) the product of the x-coordinates of the tangent points is equal to r

^{2}... ie ps=r

^{2}

b) and the product of the y-coordinates of the two tangent points is twice r

^{2}, or qt=2r

^{2}

c) the x-intercept of the common tangent is the negative of the x-coordinate of the tangent point on the parabola.

My students like solutions with numbers as examples, so here we go... if the tangent point on the circle is at (-3,4) then the tangent line will be the equation -3x+4y=25

The x- intercept of the tangent line is at -25/3, so the value of the x-coordinate of the tangent point on the parabola is at 25/3.

The y-intercept of the tangent line is at 25/4, and so the y-coordinate of the tangent point on the parabola is 25/2.

Now that we know x and y, we can find a, since y

^{2}= ax; we can write (25/2)

^{2}= a(25/3) . Then a must equal 75/4...

Checking children, we know the slope of the tangent to y

^{2}= ax will be a/2y which is (75/4) divided by (25).... but that is 3/4, and of we know that the tangent is perpendicular to the radius going from (0,0) to (-3, 4) which agrees with m=3/4 .

The real challenge now, if you are up to it, ....Assume the circle has a radius of r (pick your favorite value), and the parabola is y

^{2}= ax (pick your favorite a) and now, find the equation of the tangent line, and the points on each where the tangent point falls.