# Finding the length of the major and minor axis of ellipse.

• yungman
In summary, using differential calculus, you can find the length of the major and minor axis by finding the maximum and minimum of the derivative of the vector.
yungman
$$\vec E\;=\; \hat x E_{x0}\cos(\omega {t} -kz)\;+\;\hat y E_{y0}\cos(\omega{t}-kz+\delta)$$

For z=0, this is a vector that trace out an ellipse with time t.

I want to

1) to verify that using the definition of differential calculus, we can find the length of the major and minor axis by finding the maximum and minimum of the the derivative of the $\vec E$. This is by taking the derivative of $\vec E$ and let the derivative equal to zero to get the minimum and maximum value.

2) Find a way to get the maximum and minimum. I don't know how to do this. This is my work:

$$\frac {\partial \vec E}{\partial t}\;=\; -\hat x \omega E_{x0}\sin(\omega {t} -kz)\;-\;\hat y E_{y0}\sin(\omega{t}-kz+\delta)$$

For $\frac {\partial \vec E}{\partial t}\;=\;0$, both components of x and y has to be zero. I don't know how to do this. Please help.

thanks

Alan

yungman said:
$$\vec E\;=\; \hat x E_{x0}\cos(\omega {t} -kz)\;+\;\hat y E_{y0}\cos(\omega{t}-kz+\delta)$$

For z=0, this is a vector that trace out an ellipse with time t.

I want to

1) to verify that using the definition of differential calculus, we can find the length of the major and minor axis by finding the maximum and minimum of the the derivative of the $\vec E$. This is by taking the derivative of $\vec E$ and let the derivative equal to zero to get the minimum and maximum value.

2) Find a way to get the maximum and minimum. I don't know how to do this. This is my work:

$$\frac {\partial \vec E}{\partial t}\;=\; -\hat x \omega E_{x0}\sin(\omega {t} -kz)\;-\;\hat y E_{y0}\sin(\omega{t}-kz+\delta)$$

For $\frac {\partial \vec E}{\partial t}\;=\;0$, both components of x and y has to be zero. I don't know how to do this. Please help.

thanks

Alan

Hi Alan!

It's not ##\partial \vec E \over \partial t## that has to be zero.

What you are looking for are the 4 points where ##\vec E## is perpendicular to its tangent vector.
Those points correspond to the long axis and the short axis.
The tangent vector is ##\partial \vec E \over \partial t##.
Their dot product has to be zero.
So:

$$\vec E \cdot {\partial \vec E \over \partial t} = 0$$

$$(\hat x E_{x_0}\cos(\omega {t} -kz)+\hat y E_{y_0}\cos(\omega{t}-kz+\delta)) \cdot (-\hat x \omega E_{x_0}\sin(\omega {t} -kz)-\hat y \omega E_{y_0}\sin(\omega{t}-kz+\delta)) = 0$$

$$- \omega E_{x_0}^2 \sin(\omega {t} -kz) \cos(\omega {t} -kz) - \omega E_{y_0}^2 \sin(\omega{t}-kz+\delta) \cos(\omega{t}-kz+\delta) = 0$$

Since ##\sin(2x) = 2 \sin x \cos x##, we get:

$$E_{x_0}^2 \sin(2(\omega {t} -kz)) + E_{y_0}^2 \sin(2(\omega{t}-kz)+2\delta)) = 0$$

You can further simplify this with more trig formulas, yielding a solution for t, and after that, the 4 extreme points you are looking for.

Thank you for the help. Now it make sense. It's been a while I worked with vectors and their tangents. The moment you mentioned this, now I remember.

Thanks again

Alan

I want to verify the rest of the steps:

For z=0

$$E^2_{x0}\sin(2\omega t) + E^2_{y0}\sin(2\omega t+2\delta)=0\;\Rightarrow \sin(2\omega t)\;=\;-\frac{E^2_{y0}}{E^2_{x0}}\sin(2\omega t +2\delta)$$

$$\sin(A+B)=\sin A \cos B\;+\;\cos A \sin B\;\Rightarrow$$
$$\frac{E^2_{y0}}{E^2_{x0}}\;=\;-\frac{\sin 2\omega t}{\sin 2\omega t \cos 2 \delta + \cos 2\omega t \sin 2\delta}\;=\; -\frac{\tan 2\omega t}{\tan 2\omega t \cos 2 \delta + \sin 2\delta}$$
$$\Rightarrow\; \tan 2\omega t \;=\;-\frac{E^2_{y0}}{E^2_{x0}} tan 2\omega t \cos 2\delta \;-\;\frac{E^2_{y0}}{E^2_{x0}} \sin 2\delta\;\Rightarrow\; \tan 2\omega t (1+\frac{E^2_{y0}}{E^2_{x0}} \cos 2\delta)\;=\; -\frac{E^2_{y0}}{E^2_{x0}} \sin 2\delta$$
$$\Rightarrow \tan 2\omega t\;=\; -\frac { \sin 2 \delta}{\frac{E^2_{x0}}{E^2_{y0}}+\cos 2\delta}$$
Therefore:
$$\omega t\;=\;\frac {1}{2} \tan^-1 \left(-\frac {\sin 2 \delta}{\frac{E^2_{x0}}{E^2_{y0}}+\cos 2\delta}\right)$$

Since $\omega t$ is in radian, so I just measure the angle from x-axis. I should get two angles.

Looks good. ;)

Due to the use of ##\tan##, you will get one angle between ##- \frac \pi 4## and ##+ \frac \pi 4##.
From periodicity, the other 3 angles are a multiple of ##\frac \pi 2## from there.

If you fill in your solutions in ##\vec E## and take their lengths, you'll find the half long axis and the half short axis.

Thanks

Alan

## What is an ellipse?

An ellipse is a shape that resembles a flattened circle. It has two main axes - a major axis and a minor axis - that intersect at the center of the ellipse.

## How do you find the length of the major axis of an ellipse?

The length of the major axis can be found by measuring the longest diameter of the ellipse. This diameter passes through the center of the ellipse and its length is twice the length of the semi-major axis.

## How do you find the length of the minor axis of an ellipse?

The length of the minor axis can be found by measuring the shortest diameter of the ellipse. This diameter also passes through the center of the ellipse and its length is twice the length of the semi-minor axis.

## What is the semi-major axis of an ellipse?

The semi-major axis is half the length of the major axis. It is the distance from the center of the ellipse to the farthest point on the ellipse along the major axis.

## What is the semi-minor axis of an ellipse?

The semi-minor axis is half the length of the minor axis. It is the distance from the center of the ellipse to the farthest point on the ellipse along the minor axis.

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