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

Solution 1:

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: $$ a={b^2-d^2\over 2d}\tan\gamma. $$