How to find radius of ellipse at any point $(x_1,y_1)$.
We know semi-major axis and semi-minor axis i.e. $a$ & $b$.
center of ellipse $(x_0,y_0)$.
Somewhere I found.
$$ r = \frac{ab}{\sqrt{ a^2 \sin^2\theta + b^2 \cos^2\theta}}$$
but here $θ$ is unknown?
Please help....
On
Let the equation of the ellipse be $$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1\ \ \ \ (1)$$
Now if we set any point $P(r,\theta)$ as $\displaystyle x-x_0=r\cos\theta,y-y_0=r\sin\theta,$
we just need to the values of $x-x_0,y-y_0$ in $(1)$ and simplify
Observe that $\displaystyle\frac{x-x_0}{y-y_0}=\frac{\cos\theta}{\sin\theta}$ which demands the use of atan2
Hence $theta$ will no longer be unknown
The $\theta$ represents the angle between the $x$-axis and the line which passes the center and $(x_1,y_1)$.
In other words, the $\theta$ satisfies $$\tan\theta=\frac{y_1}{x_1}$$ if $\theta\not=\frac{\pi}{2}.$
See this figure.