## Detetmining equation of a path traced out by a point from two seperate graphs?

(at some points below I use the symbols "{" and "}" to denote an ordered pair, instead of "(" and ")" since some of the ordered pairs are functions of numbers, and would result in two ")" next to each other. Just making a note to ensure clarity)

Okay, I was trying to do this problem earlier in my notebook, and I just couldn't figure it out. Consider a circle, which we'll call circle $C$, of radius 2, centered at the origin, $O = (0, 0) \;$.

Now, consider also an ellipse, which has its first focus, which I will call:

$f_1$,

placed on circle $C$. So, this means:

$f_1 = (2, 0)$.

Also, some more details, this elipse (at its starting point) has a horizontal major axis (so then a vertical minor axis), and it's major-axis is ontop of the x-axis of course (remeber, this is just at its starting point though), and the center of the elipse is:

$c = (4, 0)$

and therefore the second focus is:

$f_2 = (6, 0)$

Remeber though, all the above is only describing the elipse's starting position. Now imagine that we have a time variable, $t$, which will help us keep track of all the movement, and might make it easier to finnally write an equation in the end, parametrically maybe. But anyway, so the position of the circle and the elipse was described at:

$t = 0$

Just a few more things about this elipse, at $t=0$, of course. Its major vertices will be called, at time $t$, $v_1(t)$ and $v_2(t)$, and are given to be (at $t = 0$):

$v_1(0) = (0, 0)$

and:

$v_2(0) = (8, 0)$

also, we will only have $t$ such that:

$0 \leq t$

Now, as $t$ increases, picture the elipse moving, such that if $f_1(t) = [a_1(t), b_1(t)]$ at time $t$, then no matter the movement, the position of $f_1$ will always satisfy:

$[a_1(t)]^2 + [b_1(t)]^2 = r^2$

Where $r= \emph{The \; radius \; of \; circle \;} C = 2$

In other words, $f_1$ stays on the path of circle C no matter what time $t$ we have.

Now, the elipse moves in such a manner that ( given that we define the second focus at time $t$ to be $f_2(t) = [a_2(t), b_2(t) ]$, and we define a line $N_c$ at time $t$, as:

$N_c(t) = (m)[x-a_1(t)] + b_1(t)$

where

$m = \frac{b_2(t) - b_1(t)}{a_2(t) - a_1(t)}$

We define this line $N_c$ also, if the above did not cover it, to always be normal to the circle C...
In other words, as the point $f_1$ moves around the circle, the elipse does not dialate or stretch, but rather it moves so that its major axis always "makes a right angle with the circle C". (more properly, it makes a 90 angle with the line tangent to the circle $C$ at $f_1$ )

And, finnaly, imagine a point that lies on the elipse. At $t=0$ this point would be at the left most vertex on the elipse, $v_1(0)$, that is. Lets denote this point as:

$\Delta$

and remember that:

$\Delta(0) = v_1(0) = (0, 0)$

Now, imagine $\Delta$ following the path of the elipse over values of $t$ so that it makes a full revolution at $t=5$, and consequently continues to do so for every multiple of 5t, but of course only speaking of its movement with respect to the ellipse alone, for now. The direction of its movement on the elipse shall always stay the same, and will start as being clock-wise, in terms of clock-wise at the elipsses position at $t=0$; so the point $\Delta$ moves upward at first an slowly curves to the right and then half way through its orbit it hits the major axis coming from the top.

Also, the elipse will move around the circle C so that is makes a full orbit (speaking as if the whole elipse where one object) first at $t = 12$ and then every multiple of $12$ afterward.

So, with all that in mind, how would one go about deriving the equation of the path that the point $\Delta$ would trace out? I know where to start, but I keep getting stuck, and am having trouble figuring it out. Any help would be much appreciated.