We have orbiting body and can calculate the orbit shape. It's ellipse with semi-major axis a and eccentricity e.
Body can move clockwise and counterclockwise on that ellipse.
How to find the point on arc in direction on movement which chord length from given point is L?
Is it possible to do the same calculations if we replace ellipse by hyperbola?
On
Some general hints:
You need an equation for the ellipse, and the coordinates of your given point $(p,q)$ on the ellipse. Then one can form the equation for the circle of radius $L$ centered at $(p,q)$ namely $(x-p)^2+(y-q)^2=L^2.$ you now have two quadratic equations in $x,y$, namely the ellipse and the circle. In general their intersection (found by solving both equations simultaneously) can be up to four different points. Then depending on which way the ellipse is oriented you have to decide which point you want. It may be the first point that a point travelling on the ellipse in your chosen direction reaches after passing through the given point $(p,q).$ But keep in mind the orbiting point will go through all the intersections found earlier, and it's up to you to decide which one you want.
Suppose you have the ellipse
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \hspace{60pt} (1)$
Let $(x_1, y_1)$ be on that ellipse. Now you want to find the point $(x,y)$ such the Euclidean distance between the two points is a given value. Therefore,
$ (x - x_1)^2 + (y - y_1)^2 = R^2 \hspace{60pt}(2) $
where $R$ is the given chord length. Equation $(2)$ is an equation of a circle whose center is $(x_1, y_1)$ and radius $R$.
We have to intersect equations $(1)$ and $(2)$, and this will result in up to $4$ points $(x,y)$, one being in a counter clockwise direction from $(x_1, y_1)$ and one in a clockwise direction from this point.
The parametric equation of the ellipse is
$(x, y) = (a \cos t, b \sin t ) \hspace{60pt}(3) $
So that $(x_1, y_1) = (a \cos t_1, b \sin t_1 ) $
where $t_1 = \text{ATAN2}( x_1 / a , y_1 / b ) $
Substitute $(3)$ into $(2)$
$ (a \cos t - x_1 )^2 + (b \sin t - y_1)^2 = R^2 $
Expanding results in
$ a^2 \cos^2 t + b^2 \sin^2 t - 2 a x_1 \cos t - 2 b \sin t y_1 + x_1^2 + y_1^2 - R^2 = 0 \hspace{60pt}(4) $
To solve equation $(4)$ use the following substitution: $ z = \tan \frac{t}{2} $
Then $z^2 + 1 = \sec^2 \frac{t}{2} = \dfrac{2}{ 1 + \cos t } $ from which it follows that
$\cos t = -1 + \dfrac{2}{z^2 + 1} = \dfrac{ 1 - z^2}{z^2 + 1 } \hspace{48pt}(5) $
Now $\sin^2 t = 1 - \cos^2 t = 1 - \dfrac{(1 - z^2)^2 }{(1 + z^2)^2 } $
From which it follows that $ \sin t = \dfrac{ 2 z }{z^2 + 1 } \hspace{48pt} (6) $
Using $(5), (6)$ equation $(4)$ becomes of the form $f(z) = 0$ where $f$ is a quartic polynomial in $z$.
Solving for its real roots $z_i$, the values of $ t_i$ is found using $t_i=2\tan^{-1} z_i$.
There will be up to four solutions, and one can determine which $t_i$ corresponds to a motion direction that is counter clockwise or clockwise.
Finally, if you replace the ellipse with a hyperbola, then you're looking at
$ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 $
whose parametric form is $(x, y) = ( a \sec t , b \tan t ) $
Plugging these into the circle equation, then multiplying through by $\cos^2 t$ results in an equation in $\cos t , \sin t , \cos 2 t, \sin 2 t $ and the same method of solution outlined above applies.