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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

2 Answers 2

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
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@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

Given a center for the ellipse at $C = (x_c, y_c)$ and a line that goes through $C$ with slope $s$ through the major axis of the ellipse, its equation is given by the zeros of $L(x,y) = y - y_c - s(x - x_c)$ and the line perpendicular to it (and also passing through $C$) is given by the zeros of $l(x,y) = s(y-y_c)+(x-x_c)$. The equation of the ellipse is given by the zeros of

$$E(x,y) = L(x,y)^2/a + l(x,y)^2/b - 1$$

Requiring that the distance between the intersections of $E$ and $L$ be $2M$ identifies $$b=M^2(1+s^2)$$ and similarly, requiring that the intersections between $E$ and $l$ be separated by $2m$ identifies $$a=m^2(1 + s^2)$$ This is demonstrated in the following SymPy session:

>>> from sympy import *
>>> a, b, x, y, m, M, x_c, y_c, s = var('a,b,x,y,m,M,x_c,y_c,s')
>>> L = (y - y_c) - s*(x - x_c)
>>> l = s*(y - y_c) + (x - x_c)
>>> idiff(L, y, x) == -1/idiff(l, y, x)  # confirm they are perpendicular
True
>>> E = L**2/a + l**2/b - 1
>>> xy = (x, y)
>>> sol = solve((E, L), *xy)
>>> pts = [Point(x, y).subs(zip(xy, p)) for p in sol]
>>> solve(pts[0].distance(pts[1]) - 2*M, b)
[M**2*(s**2 + 1)]
>>> sol = solve((E, l), *xy)
>>> pts = [Point(x,y).subs(zip(xy, p)) for p in sol]
>>> solve(pts[0].distance(pts[1]) - 2*m, a)
[m**2*(s**2 + 1)]

So the general equation of the ellipse centered at $(x_c, y_c)$ whose major axis (with radius of $M$) is on a line with slope $s$, and whose minor axis has radius of $m$, is given by the solutions of: $$\frac{((y - y_c) - s(x - x_c))^2}{m^2(1 + s^2)} + \frac{(s(y - y_c) + (x - x_c))^2}{M^2(1 + s^2)} = 1$$

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