# "Stretching" an ellipse

• Aug 6th 2009, 02:14 AM
Marril
"Stretching" an ellipse
Suppose I have an ellipse with known semi-axis lengths. Its center point is in the origin and it is rotated about the x axis in some known angle $alpha$.

I want to "stretch" this ellipse in the direction of some other angle $gamma$.
By stretching I mean:
1) Rotating the system by gamma.
2) Multiplying the new Y dependent variable (in the new coordinate system) by some constant.
3) Rotating the system backwards by gamma.

I believe the resulting shape should still be an origin centered ellipse, with new semi-axis lengths and a new rotation angle about the x-axis. So far I've failed to prove it. Anyone have an idea?
• Aug 6th 2009, 02:50 AM
alunw
You are applying three invertible linear transformations to your ellipse, so I'm sure you are right that the shape is still an ellipse. You should be able to prove this by finding the equation of the new shape. That should be routine, if messy: take a point on the new shape, apply the inverse of the three transformations you applied to the starting ellipse, now you have a point on the old ellipse, so you can plug the new point into the equation of the first ellipse.
Unless you want an explicit equation for the new shape you should be able to convince yourself the shape is still an ellipse by reasoning as follows: Any equation of the form
axx+bxy+cyy+dx+ey+f=0 is the equation of one of the following shapes : circle,ellipse,parabola,hyperbola, straight line, a pair of straight lines, or possibly 0,1 or 2 points. The substitution I am suggesting to find the new equation is linear in x and y, so you will end up with an equation of the same form. But clearly invertible linear transformations cannot turn an ellipse into any of the other possibilities apart from a circle. So applying invertible linear transformations to an ellipse you can only finish up with another ellipse or a circle.
• Aug 6th 2009, 03:37 AM
Marril

Luckily I'm not pressed to know the explicit function of the new ellipse, as the new axis lengths and rotation angle can be easily found numerically. Calculating it would be somewhat helpful, but I'm getting to a system of two parametric equations which I'm having trouble to unlock.
• Aug 6th 2009, 04:05 AM
malaygoel
Quote:

Originally Posted by Marril

but I'm getting to a system of two parametric equations which I'm having trouble to unlock.

• Aug 6th 2009, 06:58 AM
Marril
Here are the calculations I made earlier (in the attachment). For simplicity I checked for a simple initial ellipse, with axis parallel to x and y axes.
• Aug 6th 2009, 07:10 PM
malaygoel
Quote:

Originally Posted by Marril
Here are the calculations I made earlier (in the attachment). For simplicity I checked for a simple initial ellipse, with axis parallel to x and y axes.

With the notation in your document followed, I got this:

$\frac{\{ \overline{x}(\cos^2\gamma +m\sin^2 \gamma) + \overline{y}(1-m)\sin \gamma \cos \gamma\}^2}{a^2}$+ $\frac{\{ \overline{x}(1-m)\sin \gamma \cos \gamma + \overline{y}(\sin^2\gamma +m\cos^2 \gamma)\}^2}{b^2}$ $=1$

where m=1/f
• Aug 6th 2009, 10:23 PM
Marril
Thanks malaygoel! (Clapping)
Your result definitely stands for an origin centered ellipse. Carrying out the transformation any number of times will give me ellipses then.
• Sep 1st 2009, 10:08 PM
Marril
I still wish to torment my ellipse some more.

Suppose I have a general origin centered ellipse of the form $R(\theta) = \frac{a \cdot b}{\sqrt{a^2\cdot sin^2(\theta-\omega)+b^2\cdot cos^2(\theta-\omega)}}$

Now I want to stretch it using the recipe stated above (Namely rotating by some angle $\gamma$ and multiplying the y component in the new coordinate system by another constant $f$). We already arrived at the coclusion that the result would be another origin centered ellipse.
This means that the new ellipse can be written in the form $R'(\theta) = \frac{A \cdot B}{\sqrt{A^2\cdot sin^2(\theta-\Omega)+B^2\cdot cos^2(\theta-\Omega)}}$

The question is, how can I find the parameters ( $A, B, \Omega$) as functions of ( $a, b, \omega, \gamma, f$)?
• Sep 5th 2009, 08:28 AM
Marril