Points on two ellipses with identical tangent lines

1. Aug 7, 2010

ZippyDee

Hi, I'm trying to get this working for a program I'm making. I've been working on this for a while, but I can't seem to figure it out.

I have multiple rotated ellipses. Imagine you took a rubber band and stretched it around the ellipses. The rubber band would follow the curve of the outside of an ellipse until it reached a point on the ellipse whose tangent line was the same as the tangent line of another point on the next ellipse. What I need to figure out is where those points are.

For each ellipse, I know: center x, center y, semimajor axis, semiminor axis, and the amount by which it has been rotated. So any ellipse E has known variables x, y, a, b, and theta.

I don't have equations for the ellipses, all I have are those variables.

How do I go about solving this?

-Zippy Dee

2. Aug 8, 2010

Petr Mugver

I assume you can write the cartesian equation of the ellipse (if you don't, I'll explain it in the next post). This is of the form

$$A_1x_1^2+B_1y_1^2+C_1x_1y_1+D_1x_1+E_1y_1+F_1=0$$ (1)

and the same for the other ellipse:

$$A_2x_2^2+B_2y_2^2+C_2x_2y_2+D_2x_2+E_2y_2+F_2=0$$ (2)

where all the A,B,... are known numbers.

Now write another two equations:

$$2A_1x_1(x-x_1)+2B_1y_1(y-y_1)+C_1x_1(y-y_1)+C_1y_1(x-x_1)+D_1(x-x_1)+E_1(y-y_1)=0$$

and the same for the second one:

$$2A_2x_2(x-x_2)+2B_2y_2(y-y_2)+C_2x_2(y-y_2)+C_2y_2(x-x_2)+D_2(x-x_2)+E_2(y-y_2)=0$$

If you collect terms, you can write the last two equations as

$$y=G_1x+H_1$$

and for the second one

$$y=G_2x+H_2$$

Now put

$$G_1=G_2$$ (4)

and

$$H_1=H_2$$ (3)

You have to solve the system composed of equation (1), (2), (3), (4). You will get the four unknowns x_1, y_1, x_2 and y_2. Typically (but not always) you will find four 4-uples of solutions.

3. Aug 8, 2010

ZippyDee

Thank you. That makes a lot of sense! However, I am not sure how to write the Cartesian equations for the ellipses. Could you explain how to do that?

4. Aug 8, 2010

Petr Mugver

Start vriting the canonical equation of an ellipse centered at the origin and "unrotated":

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Then rotate it (clockwise) by the angle theta. This means you have to do the transformation

$$x\rightarrow x\cos\theta-y\sin\theta$$
$$y\rightarrow x\sin\theta+y\cos\theta$$

i.e.:

$$\frac{(x\cos\theta-y\sin\theta)^2}{a^2}+\frac{(x\sin\theta+y\cos\theta)^2}{b^2}=1$$

And finally, put the center of the ellipse in the right place:

$$x\rightarrow x+x_C$$
$$y\rightarrow y+y_C$$

that is,

$$\frac{[(x+x_C)\cos\theta-(y+y_C)\sin\theta]^2}{a^2}+\frac{[(x+x_C)\sin\theta+(y+y_C)\cos\theta]^2}{b^2}=1$$

(I won't expand this for you! )

Other questions?