Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've tried to detail my question using the image shown in this post. ellipse normals.

Consider an ellipse with 5 parameters $(x_C, y_C, a, b, \psi)$ where $(x_C, y_C)$ is the center of the ellipse, $a$ and $b$ are the semi-major and semi-minor radii and $\psi$ is the orientation of the ellipse.

Now consider a point $(x,y)$ on the circumference of the ellipse. The normal at this point on the circumference of the ellipse intersects the major axis at a point $(x_D, y_D)$. This normal makes an angle $/phi$ with the major axis. However, the angle subtended by this point at the center of the ellipse is $\theta$. For a circle, $\theta = \phi$ for all points on its circumference because the normal at the circle is the radial angle subtended by the point on the circumference.

Is there a relationship between the angles $\theta$ and $\phi$ for an ellipse.

For some context, I am trying to "extract" points from the circumference of an ellipse given its parameters $(x_C, y_C, a, b, \psi)$. For such an ellipse, I start from $(x_C, y_C$) and with angle $\theta = 0^\circ$ and I start sweeping until $360^\circ$. Using the equation $\left[\begin{array}{c} x \\ y\end{array}\right] = \left[\begin{array}{c c} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array}\right] \left[\begin{array}{c} a\cos(\psi) \\ b\sin(\psi) \end{array}\right]$, I get the $(x,y)$ location of the point that is supposed to be on the ellipse circumference. I then look up this location in a list of "edge" points. Along with this list of edge points, I also have gradient angle information for each edge point. This corresponds to the angle $\phi$.

Here is the crux of the question, for a circle, I am confident that the edge point lies on the circumference of the circle if $|\theta - \phi| < \text{threshold}$. But, for an ellipse, how do I get a similar relationship ?

share|improve this question
    
Does this help you? math.stackexchange.com/questions/171936/… –  Jorge Fernández Jul 25 '12 at 19:01
    
what does that equation refer to? The slope of the tangent line? If, so then the slope of the normal is just -1/slope of tangent right ? –  Mustafa Jul 25 '12 at 19:12
    
yes, thats right –  Jorge Fernández Jul 25 '12 at 19:14
    
Could you point me to how you derived that equation? –  Mustafa Jul 25 '12 at 20:30
    
Do you know calculus? –  Jorge Fernández Jul 25 '12 at 20:44

3 Answers 3

Take $P = (p_x, p_y)$ on the circumference to be functions of angle, $\theta$.

Now use the Cartesian equation of an ellipse:

$x^2/a + y^2/b = 1$

Differentiate w.r.t. $\theta$:

$2 \frac{dx}{d\theta} x / a + 2 \frac{dy}{d\theta} y / b = 0$

Cancel the 2:

$\frac{dx}{d\theta} x / a + \frac{dy}{d\theta} y / b = 0$

As we want the gradient at a particular point or angle, we know the value of $(x, y) = (p_x, p_y)$, so:

$\frac{dx}{d\theta} p_x / a + \frac{dy}{d\theta} p_y / b = 0$

Divide through by $\frac{dx}{d\theta}$ if you want $\frac{dy}{dx}$:

$p_x / a + \frac{dy}{dx} p_y / b = 0$

To get the line equation of the tangent, use the direction from this line equation plus a point on the line (e.g. where the tangent meets the ellipse):

$L = (p_x, p_y) + \mu (-b\ p_y, a\ p_x)$

share|improve this answer

First solve the problem for a horizontal ellipse at the origin, and then transform the result to the desired coordinates.

If any curve is given in polar coordinates as $r(\theta)$ then the angle between the normal and the position vector is

$$\tan \psi= - \frac{r'(\theta)}{r(\theta)} $$

Why? Look at the picture below. Imagine a curve (green) defined in polar coordinates. The tangent unit vector is decomposed into ${\rm d}r$ and $r {\rm d}\theta$ components.

Cam

The angle $\psi$ from the radius to the normal is $$\tan\psi =- \frac{{\rm d}r}{r {\rm d}\theta}$$

The negative sign is to indicate that it moves in the opposite direction from $\theta$.

From this you can work out the rest you need.

Horizontal Ellipse:

$$r(\theta) = \frac{a b}{\sqrt{a^2-(a^2-b^2)\cos^2\theta}} $$

$$\tan \psi =\frac{(a^2-b^2)\sin\theta\cos\theta}{a^2 - (a^2-b^2)\cos^2\theta} $$

share|improve this answer

With the center at $(0,0)$ and $\psi=0$, the equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$$ So $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$ and therefore $$\frac{dy}{dx}= -{\frac{xb^2}{ya^2}}\text{.}$$ The slope of the normal line is the negative reciprocal of this, so $$\tan(\phi)=\frac{ya^2}{xb^2}\text{.}$$ Meanwhile, $$\tan(\theta)=\frac{y}{x}\text{.}$$ So, eliminating $\frac{y}{x}$ from these two equations and clearing denominators, the relationship between $\phi$ and $\theta$ is: $$b^2\tan(\phi)=a^2\tan(\theta)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.