# Normal to Ellipse and Angle at Major Axis

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

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 ?

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Does this help you? math.stackexchange.com/questions/171936/… – Carry on Smiling 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 – Carry on Smiling 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? – Carry on Smiling Jul 25 '12 at 20:44

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: $$b^2\tan(\phi)=a^2\tan(\theta)$$

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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.

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:

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

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

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

Now use the Cartesian equation of an ellipse:

$\frac{x^2}{a} +\frac{y^2}{b} = 1$

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

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

Cancel the 2:

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

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

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

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

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

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):

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

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