What is happening in the picture I came across the picture below through random means.

What is being demonstrated? All I could think of is maybe the center of the triangle is moving back and forth between the focii of the ellipse, but even if that's true (which it may or may not be - it's purely conjecture on my part), there's clearly more going on here.
 A: We need to prove the following: when two vertices of a fixed triangle slide along two arms of a fixed angle, the locus described by the third vertex is an ellipse.
In the figure  all the letters excepting $t$ and $P(x,y)$ are data of the problem; the angle $t$ determines the position of the vertex $P(x,y)$ so we choose $t$ as a parameter to be eliminated in order to find out the searched locus.

We have $$OB=\frac {c\sin (\theta+t)}{\sin\theta}$$ then
$$x= \frac {c\sin (\theta+t)}{\sin\theta} \,–\, a\cos(\beta+t)$$
$$y=a\sin(\beta+t)$$
i.e. $$x=A\cos t-B\sin t$$
$$y=C\cos t+D\sin t$$
Where $$A=c-a\cos\beta$$$$B=c\cot \theta +a\sin\beta$$
$$C=a\sin\beta$$ $$D=a\cos\beta$$
Solving for $\sin t$ and $\cos t$ and because of $$\sin^2 t+\cos^2 t=1$$ it follows 
$\det^2\begin{pmatrix}x&-B\\y&D\\ \end{pmatrix}$+$\,\det^2\begin{pmatrix}A&x\\C&y\\ \end{pmatrix}=$$\,\det^2\begin{pmatrix}A&-B\\C&D\\ \end{pmatrix}$
Thus $$(Dx+By)^2+(Ay-Cx)^2=(AD+BC)^2$$i.e.$$(C^2+D^2)x^2+(A^2+B^2)y^2+2(BD-AC)xy=(AD+BC)^2$$ This is a quadratic form of two real variables in which coefficients of $x^2$ and $y^2$ are both positive and distinct. This shows that the locus is an ellipse.

A: You are given a circle with two diameters that are not perpendicular to each other. Two points go back and forth along the diameters. A third point creates a triangle. As the two points move along the diameters, the third point of the triangle traces an ellipse. 
