# Math Help - Help with a non-canonical ellipse needed

1. ## Help with a non-canonical ellipse needed

I am accustomed to working with ellipses in canonical form. But I've got one that is definitely not in canonical form, and I am at a complete loss as to how to analyze it.

The original equations are these:

$
x=3cos(t)
$

$
y=2sin(t+pi/4)
$

I need to figure out the long and short radiuses and the inclination angle of the long axis.

I am completely at an utter loss as to how to do this. I've tried to work with the equations every way I can think of, but I can't figure it out.

Any help would be welcome!

2. Originally Posted by Lancet
I am accustomed to working with ellipses in canonical form. But I've got one that is definitely not in canonical form, and I am at a complete loss as to how to analyze it.

The original equations are these:

$
x=3\cos(t)
$

$
y=2\sin(t+\pi/4)
$

I need to figure out the long and short radiuses and the inclination angle of the long axis.
You could start like this:

$x=3\cos(t),$
$y=2\sin(t+\frac\pi4) = 2(\sin t\cos\frac\pi4 + \cos t\sin\frac\pi4) = \sqrt2(\sin t+\cos t).$

Therefore $\left(\frac x3\right)^2 + \bigl(\frac y{\sqrt2} - \frac x3\bigr)^2 = 1.$

Now you have the cartesian equation of the ellipse, and you can use standard methods to get the orientation and the lengths of the axes.

3. This is the first time I've ever had to deal with a rotated ellipse. I tried looking up information on how to interpret the form of equation you helpfully supplied, but I'm coming up blank. The closest I came was Q21 here, but I saw no clear way I could use that to interpret what you provided.

Can you (or someone) point me in the right direction?

4. Originally Posted by Lancet
This is the first time I've ever had to deal with a rotated ellipse. I tried looking up information on how to interpret the form of equation you helpfully supplied, but I'm coming up blank. The closest I came was Q21 here, but I saw no clear way I could use that to interpret what you provided.

Can you (or someone) point me in the right direction?
In brief. First write it out as Opalg said. That will give you the form
$ax^2 + bxy + cy^2 = d$

In general this ellipse will be at an angle. (The non-zero xy term tells you this.) So we need to get rid of the xy term by a change in coordinates. Since the ellipse is rotated, let's choose a rotated set of coordinates:
$x = x'cos( \theta ) + y' sin( \theta )$

$y = -x' sin( \theta ) + y' cos( \theta )$

Mind you we don't yet have a value for the angle $\theta$ yet, so you'll need to do a fair amount of algebra. After a while you will get the new equation to be of the form
$a( \theta )x'^2 + b( \theta ) x'y' + c( \theta )y'^2 = d$

For this to be useful, ie to get the equation into an "unrotated" form, the coefficient of x'y' must be zero. So set $b( \theta ) = 0$ and solve for $\theta$. Then put that angle back into your x', y' equation and simplify. That will get you an ellipse of the standard form after a bit more algebra.

This is a long and nasty problem in general. To do yours I had to use numerical methods the whole way. But the answer did come out eventually.

-Dan

Oh yeah I forgot to mention. It can get worse. Notice that the original form (the xy format) of your ellipse shows that the ellipse is centered on the origin. (The terms are $ax^2$ and $cy^2$ as opposed to $ax^2 + ex$ and $cy^2 + fy$.) If the ellipse is not centered on the origin then you need to perform a coordinate transform to translate the center to the origin before the rotation trick.

Fortunately they don't assign many of these for you to do. And I've never seen one on an exam.

5. That sounds dismayingly daunting. I'm grateful for your attempt, but I admit I'm not hopeful of being able to pull it off.

By any chance is it easier to work from the original parametric equations?

6. Originally Posted by Lancet
That sounds dismayingly daunting. I'm grateful for your attempt, but I admit I'm not hopeful of being able to pull it off.

By any chance is it easier to work from the original parametric equations?
You're quite right, this problem is not rigged to have a neat solution. Here's another method that seems a bit less daunting.

What we know so far is that we can write $x = 3\cos t,\ y = \sqrt2(\cos t+\sin t).$ We also know, from topsquark's comment, that the ellipse is centred at the origin. This means that the length and direction of the axes can be determined by looking at the points on the ellipse that are closest to, or furthest from, the origin.

The square of the distance of (x,y) from the origin is

$x^2+y^2 = 9\cos^2t+2(\cos t+\sin t)^2 = 11\cos^2t+4\cos t\sin t +2\sin^2t.$

Differentiate that (and simplify the result just a bit) to find that the turning points occur when $2\sin^2t +9\cos t\sin t -2\cos^2t=0.$ Now divide through by $\cos^2t$ to get a quadratic equation in $\tan t$. Unfortunately it doesn't have a nice solution (unless I made an error in calculation, there's a $\sqrt{97}$ in the answer), but I think that's the best that can be done.

7. Originally Posted by Opalg
This means that the length and direction of the axes can be determined by looking at the points on the ellipse that are closest to, or furthest from, the origin.
Good call! That never occurred to me in the entire time I've been doing these.

-Dan