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?
What is the equation of an ellipse that is not aligned with the axis?
-
1In 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: http://en.wikipedia.org/wiki/Rotation_matrix. – 2012-02-11
3 Answers
Let's suppose it was semi-major axis $4k$ and semi-minor 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
$17\,{y}^{2}-30\,x\,y+17\,{x}^{2} - 32\,{k}^{2} = 0.$
-
8@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 $x^{\prime }=\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y$ $y^{\prime }=\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}x.$ – 2012-02-11
Let the center of the ellipse be at $C = (x_c, y_c)$. Let the major axis be the line that passes through $C$ with a slope of $s$; points on that line are given by the zeros of $L(x,y) = y - y_c - s(x - x_c)$. Let the minor axis be the line perpendicular to $L$ (and also passing through $C$); points on that line are given by the zeros of $l(x,y) = s(y-y_c)+(x-x_c)$. The ellipse is then defined 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: $\frac{((y - y_c) - s(x - x_c))^2}{m^2(1 + s^2)} + \frac{(s(y - y_c) + (x - x_c))^2}{M^2(1 + s^2)} = 1$
Sorry I didn’t see this one much earlier. Your ellipse has its major axis aligned with (I’m assuming) the line $y=x$, in other words it has symmetry $x\leftrightarrow y$. This means that it has an equation of the shape $ a(x^2+y^2) + bxy+c(x+y)=1\,, $ and since you’ve specified that the center is at the origin, you also need $c=0$. Thus your equation will simply be $a(x^2+y^2)+bxy=1$. You say that the semimajor axis is to be $4d$ and the semiminor axis is to be $d$ in length. So the points you know to be on the ellipse are $(\frac{4d}{\sqrt2},\frac{4d}{\sqrt2})=(2\sqrt2d,2\sqrt2d)$ at the end of the major axis and $(-\frac d{\sqrt2},\frac d{\sqrt2})$ on the minor axis.
Substituting these values into the equation and solving for $a$ and $b$, you get $ a=\frac{17}{32d^2}\,,\quad b=-\frac{15}{16d^2}\,, $ at least if I have not made a computational error. And there’s the equation for your ellipse.