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....


2 Answers 2


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.


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

  • $\begingroup$ after solving this I have r=ab is this correct? $\endgroup$
    – sandy
    Dec 31, 2013 at 10:01
  • 1
    $\begingroup$ @sandy, no we have $$\frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2}=1\implies r^2=?$$ $\endgroup$ Dec 31, 2013 at 10:05

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