How to eliminate angle in a Glissette equation of carried point of a line sliding along two lines not at right angles

Question

Glissettes are the curves traced out by a point carried by a curve, which is made to slide between given points or curves. My problem specifically include a line which slides between two fixed lines (which are not necessarily at right angles), and a carried point in the line (not necessarily at the center).

I found some text in Notes on Roulettes and Glissettes : William Henry Besant (pages 51–52), which solve this for two lines at right angles, and adds a note how to do it for two lines which are not at right angles, but doesn't solve it completely.

For simplicity, assume one of the fixed lines is the x-axis, and it intersects the other fixed line at the origin, and the angle between them is $\alpha$. If the moving line has a length $d_1+d_2$, and the carried point is at distances $d_1,d_2$ from the line's endpoints, the Glissette points satisfy the following for any angle $\theta$:

$$ y = d_2 \cos (\theta + \alpha) ,\quad x = \frac{d_1 \cos \theta}{\sin \alpha} + \frac{y}{\tan \alpha}. $$

How do I eliminate the angle $\theta$ from the equation? The end curve is an ellipse but I can't calculate its exact equation. I would like to have an equation that depends only on $x,y$ without $\theta$, where $d_1,d_2,\alpha$ are known and constant, $f(x,y)=0$.

Appreciate your help!

Answer 1

The equation $f(x, y)=0$ is as follows: $$\sin{\alpha}^{2}d_{2}^{2}.x^{2}-2\cos(\alpha)\sin(\alpha)(d_{1}+d_{2})x y+\Bigl(d_{1}^{2}+d_{2}(2d_{1}+d_{2}) \cos(\alpha)^{2}\Bigr) y^{2}-d_{1}^{2}d_{2}^{2} \sin(\alpha)^{2}=0.$$