# 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?