## Monday, 16 November 2009

### More on the Common Tangent Problem

I mentioned in my last blog that I had been exploring the relationships around the problem of a common tangent to a circle and a parabola with a vertex at the center of the circle. To make notation easier, I have assumed that the center and vertex are at the origin and the axis of symmetry is along the x-axis. I came across a couple of unexpected (to me) relationships, and so I thought I would present them in the form of a problem, with the answer further below.

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 y2 = ax. If the tangent contains the point (p,q) on the circle; find the coordinates of the tangent point (s,t) on the parabola?

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 x2+y2=r2 at (p,q) and a parabola, y2= ax at (s,t) the product of the x-coordinates of the tangent points is equal to r2... ie ps=r2

b) and the product of the y-coordinates of the two tangent points is twice r2, or qt=2r2

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 y2 = 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 y2 = 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 y2 = ax (pick your favorite a) and now, find the equation of the tangent line, and the points on each where the tangent point falls.