# Find angle at given points in Ellipse

I can find, how to check if given point(x, y) exists in Ellipse or not.

Now, I want to find given point(x,y) exists at which angle in Ellipse.

• you are saying a given point is on the ellipse? – mathlove Dec 19 '13 at 15:13
• It's unclear what "which angle in Ellipse" you mean. @mathlove has addressed a central angle between a ray from center (0,0) to point (x,y) on the ellipse and the semi-major axis/ray from center to (a,0). Often we are interested in angles made by lines with the ellipse itself, e.g. a ray from one focus reflecting off the ellipse into the other focus. – hardmath Dec 19 '13 at 15:33
• @mathlove , Yes, I know the point(x,y) and in ellipse. I want to know that point is at which angel in ellipse. In other words, I know point(x,y), center (cx, cy) and radii (rx, ry) of Ellispe, I have to find angle. Thanks – Vikram Dec 20 '13 at 5:07
• @Vikram: I'm confused with your comment. If you mean the point is in ellipse, and if you want to know that point is at which angle toward $x$-axis, then you just need to solve $\tan \theta=y/x$? – mathlove Dec 20 '13 at 5:14
• I want to find angle to check if given point(x,y) exists between 2 angles or not. For example, if given point is between $15^o$ and $45^o$ or not? Thanks – Vikram Dec 20 '13 at 5:20

You should know an ellipse can be represented as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a\gt b\gt 0)$. Hence, if you are saying a given point $(x,y)$ is on the ellipse, we have the following representation : $$x=a\cos\theta, y=b\sin\theta\ \ (0\le\theta \lt 2\pi).$$

Hence, if you know $(x,y)$, then you can calculate the $\theta$, which represents the angle of the point.

• Thank you for answer. I am already using this formular to check, if point is in Ellipse or not. Now, I want to check , if that point(x,y) exist at which angle of Ellipse. Thanks – Vikram Dec 20 '13 at 5:13

I got the formula to find angle towards $x$-axis at given point $(x, y)$ in Ellipse:

$\theta = \arctan2(cx - x, cy -y)\times \frac{180^\circ}{\pi} + 90^\circ$

It gives you angle from $0^\circ$ to $270^\circ$ .

To find remaining angles from $271^\circ$ to $360^\circ$, we need to do following calculation:

If $\theta <0^\circ$ then $\theta = 360^\circ + \theta$.

Thanks :)

Taking the answer of mathlove $x=acosθ, y=bsinθ$ you can derive using that $tan:=sin/cos$ this formula:

$y = b \ cosθ \ tanθ;$ with $cosθ =$ $x \over a;$

=> $y =$ $b \over a$ $x \ tanθ;$