# What is the equation of an ellipse that is not aligned with the axis?

I have the an ellipse with its semi-minor axis length $x$, and semi major axis $4x$. However, it is oriented $45$ degrees from the axis (but is still centred at the origin). I want to do some work with such a shape, but don't know how to express it algebraically. What is the equation for this ellipse?

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$x$ is not particularly helpful as a parameter if you also intend to use it as an equation variable –  Henry Feb 11 '12 at 21:35
In general, compute what you already know (say, equation of the ellipse aligned with the $x$-axis) and then apply rotation by the desired angle. Look here: en.wikipedia.org/wiki/Rotation_matrix. –  William Feb 11 '12 at 22:01

Let's suppose it was semi major axis $4k$ and semi major axis $k$ to avoid confusion.

Aligned with the axes it would be

$$\frac{x^2}{(4k)^2}+\frac{y^2}{k^2}=1$$

but you want this rotated, so replace $x$ by $\frac{x+y}{\sqrt 2}$ and $y$ by $\frac{y-x}{\sqrt 2}$ to get

$$\frac{(x+y)^2}{32k^2}+\frac{(y-x)^2}{2k^2}=1$$ which you can also write as

$$17\,{y}^{2}-30\,x\,y+17\,{x}^{2} - 32\,{k}^{2} = 0.$$

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Thanks for the answer. You're right, x was a stupid choice in variable names. How does replacing $x$ by $\frac{x+y}{\sqrt(2)}$ account for the 45 degree shift, if you don't mind me asking? :) –  deod Feb 11 '12 at 21:54
Instead of the axes being aligned along $x=0$ and $y=0$ they are aligned along the perpendicular $x+y=0$ and $y-x=0$, $45^\circ$ to the original axes. The $\sqrt 2$ is just a scaling factor. –  Henry Feb 11 '12 at 22:11
@deod: The relation of the coordinates $(x,y)$ of a point in a coordinate system and the coordinates $(x^{\prime },y^{\prime })$ of the same point in a rotated coordinated system (with the positive $x^{\prime }$- axis making an angle $\theta$ with the positive $x$ - axe) is $$x^{\prime }=x\cos \theta +y\sin \theta$$ $$y^{\prime }=y\cos \theta -x\sin \theta .$$ For $\theta =45{{}^\circ}$, $\cos \theta =\sin \theta =1/\sqrt{2}$. Then $$x^{\prime }=\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y$$ $$y^{\prime }=\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}x.$$ –  Américo Tavares Feb 11 '12 at 22:18