Problem in defining a trigonometric equation (ellipse) I have an updated problem of my question from:
Problem in defining a trigonometric equation
@David K gave me a very nice solution here. But now the problem is that since I do not have a circle but a ellipse the radius r is not perpendicular to the straight line with slope $\alpha$ anymore.
So I somehow don't know how to get to a solution because I think something is missing.
Given:

*

*$r_f$

*$\tau$

*major axis $b$

*slope $\alpha$ of straight line

*$x_\phi$,$y_\phi$
Target:

*

*Find value of minor axis $a$ with given values


I can solve this for a circle with radius r (see link from my last question) and I then thought maybe it is possible to transform the circle to a ellipse but since $a$ is unknown this seem not to work...
Here what I tried so far but I'm unable to continue because I cannot define $\beta$.

(1) $x_1=x_\phi-y_\phi cot(\tau)$
(2) $\delta = \frac{\pi}{2}-\tau$
(3) $\frac{x_1}{sin(\delta+\beta)}=\frac{r}{sin(\tau)}$
Putting (1) and (2) in (3) gives me:
--> $r=\frac{sin(\tau) (x_\phi-y_\phi cot(\tau))}{sin(\frac{\pi}{2}-\tau+\beta)}$
with unkown $\beta$
(4) $x_2=x_\phi+y_\phi tan(\beta)$
(5) $\Psi = \frac{\pi}{2}-\beta$
(6) $\frac{r_f+a}{sin(\Psi)}=\frac{x_2}{sin(\delta+\beta)}$
Putting (4), (5) and (2) in (6) gives me:
$a=(\frac{x(\phi)+y(\phi)tan(\beta)}{sin(\frac{\pi}{2}-\tau+\beta)}-\frac{r_f}{sin(\frac{\pi}{2}-\beta)})sin(\frac{\pi}{2}-\beta)$
but still $\beta$ and $x_\phi$ is unknown... Also I think $b$ can help me to find a solution for $a$, since b is known but I don't know how I can use $b$.
// edit 2021-12-07:
What about if $x_\phi$,$y_\phi$ is unkown?
// edit 2021-12-08:
After discussion with @Intelligenti pauca:

*

*Line with slope $\alpha$ and y-intersection $t$ is given (fix)

*minor axis $a$ and point of tangency $x_\phi$,$y_\phi$ need to be found.

Best regards
mk3
 A: Define the following vectors:
$$
\vec{r}_f=r_f(\cos\tau,\sin\tau),\quad
\vec{a}=-a(\cos\tau,\sin\tau),\quad
\vec{b}=b(\sin\tau,-\cos\tau),\quad
\vec{P}=(x_\varphi,y_\varphi).
$$
As $\vec P$ is a point on the ellipse we have:
$$
\vec{P}=\vec{r}_f-\vec{a}+\vec{a}\cos t+\vec{b}\sin t,
$$
which are two equations in the two unknowns $a$ and $t$. Solving them you can get $a$. Note that the value of $\alpha$ is not needed.
EDIT.
Here's a more geometric approach, exploiting our knowledge of angle $\alpha$:

*

*write the equation of the tangent line passing through point $F$, at the end of $\vec{r}_f$;


*write the equation of the tangent line passing through point $P$, at the end of $\vec{y}_\varphi$;


*find point $T$, intersection of the two tangents described above;


*find point $M$, midpoint of $FP$;


*line $TM$ passes through the center $C$ of the ellipse, which can then be found;


*once you have $C$ finding $a$ is straightforward.
Note that in this case the value of $b$ was not used in the solution.

EDIT 2.
Here's a solution for the edited question (see figure below). Now we are given an endpoint $F$ of semi-axis $FC$, the line $OF$ on which that semi-axis lies, a tangent line $PT$ and the length of the other semi-axis $b$. We want to find $a=FC$ and the position of tangency point $P$.
Let's draw the tangent $FT$ through $F$ (perpendicular to $OF$) and let $T$ be its intersection with the other tangent $PT$. Length $d=FT$ can be computed from the data, as well as the angle $\gamma$ between the tangents. Let's also set $p=PT$, a quantity to be found, and $\delta=\angle CTF$.
As midpoint $M$ of $FP$ lies on $CT$, then the fourth vertex $Q$ of parallelogram $FTPQ$ also lies on $CT$. The sine rule applied to triangle $FTQ$ gives $p:\sin\delta=d:\sin(\pi-\gamma-\delta)$, that is:
$$
\tan\delta={p\sin\gamma\over d-p\cos\gamma}.
$$
From triangle $TFC$ we then obtain:
$$
a=d\tan\delta={pd\sin\gamma\over d-p\cos\gamma}.
$$
On the other hand, $P$ is a point on the ellipse. If $H$ is the projection of $P$ on $FT$ we have then:
$$
\left({FC-PH\over FC}\right)^2+
\left({FT+TH\over BC}\right)^2=1,
$$
that is:
$$
\left({a-p\sin\gamma\over a}\right)^2+
\left({d+p\cos\gamma\over b}\right)^2=1.
$$
Inserting here the above result for $a$ we obtain a quadratic equation for $p$, which can be solved to yield:
$$
p\cos\gamma={d(b^2-d^2)\over b^2+d^2}.
$$
Finally, we can plug this into the expression for $a$, obtaining:
$$
a={b^2-d^2\over 2d}\tan\gamma.
$$

