How would I write the equation for a tilted ellipse?

Question

Stupid math teacher assigned us a last minute project during dead week. I'm graphing an image, but I"m stuck on a tilted ellipse that I don't know how to graph, and as I've already turned in my math book and the last time we covered this subject was when I was a freshman I can't figure out how to tilt this ellipse :( Please help!

(x-10)^2/5.5^2 + (y+20)^2/2^2 = 1

is the equation of the ellipse I"m trying to tilt 45 degrees. What equation would I get? I also need to convert it into parametric values :( But assuming the equation's still similar this shouldn't be a problem

Answer 1

I assume you want to rotate about ellipse centre and not (0,0).



….......



If you rotate the point (x,y) by an angle t counter-clockwise about the origin the new coordinates of the point are X = xcos(t)−ysin(t), Y = xsin(t)+ycos(t).



Inverting these give x = Xcos(t)+Ysin(t), y = −Xsin(t)+Ycos(t)



For an angle of 45°, cos(t)=sin(t)=1/√2 so we have x=(X+Y)/√2, y=(−X+Y)/√2 … (i)



…......



Translate axes by (10,−20) so ellipse equation becomes x²/5.5² + y²/2² = 1 … (ii)



Using (i) gives (X+Y)²/60.5 + (−X+Y)²/8 = 1 → 137X² − 210XY+137Y² = 968 … (iii)



(iii) is the equation of the rotated ellipse relative to the centre.



Reversing translation : 137(X−10)² − 210(X−10)(Y+20)+137(Y+20)² = 968



This is equation of rotated ellipse relative to original axes.





A parametric form for (ii) is x=5.5cos(θ), y=2sin(θ) for 0≤θ≤2π



Using (i) this converts to X+Y = √2*5.5*cos(θ), −X+Y = √2*2*sin(θ)



∴ X = (1/√2)( 5.5cos(θ)−2sin(θ) ), Y = (1/√2)( 5.5cos(θ)+2sin(θ) ) for 0≤θ≤2π



These are parametric equations for rotated ellipse relative to centre.



Reversing translation :



X = 10 + (1/√2)( 5.5cos(θ)−2sin(θ) ), Y = −20 + (1/√2)( 5.5cos(θ)+2sin(θ) )



These parametric equations describe rotated ellipse relative to original axes.

Answer 2

Rotated Ellipse Equation

Answer 3

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RE:

How would I write the equation for a tilted ellipse?

Stupid math teacher assigned us a last minute project during dead week. I'm graphing an image, but I"m stuck on a tilted ellipse that I don't know how to graph, and as I've already turned in my math book and the last time we covered this subject was when I was a freshman I...

Answer 4

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Using the parametrization for the ellipse (x = acost, y = bsint), we can express the curve with the vector-valued function r(t) = [x(t), y(t)]. We can express a counterclockwise rotation by an angle θ with the standard matrix below: |cosθ ... -sinθ| |sinθ ... cosθ|. Sorry for the sloppy notation. The "..." are just there for spacing. Let A denote the standard matrix above. Then we need to compute Ar: Ar = [acos(t)cos(θ) - bsin(t)sin(θ), acos(t)sin(θ) + bsin(t)cos(θ)]. We can split this into the parametric eqns: x(t) = acos(t)cos(θ) - bsin(t)sin(θ) y(t) = acos(t)sin(θ) + bsin(t)cos(θ) Addition: I see - -'s result matches mine. Converting to polar and then applying the rotation seems like a solid approach because it's easier to work with rotations in polar. Addition: The matrix A provides us with a tool to map the points on our ellipse to new points generated by a rotation of θ. We often write T(x) = Ax, using linear transformation notation to describe the mapping. The transformation T takes vector x and maps it to the vector obtained by performing the multiplication Ax. There is theorem in Linear Algebra that tells you how A is determined: Let e_1, e_2, ... , e_n be the columns of the identity matrix. Then the columns of the matrix A are T(e_1), T(e_2), ... , T(e_n). In our problem we are dealing with two dimensions so n = 2. In this case e_1 = [1, 0] and e_2 = [0, 1] (technically I should write these vertically). T(e_1) is the new vector generated after rotating [1, 0] by θ ccw (similar definition for T(e_2)). This gives T(e_1) = [cosθ, sinθ] and T(e_2) = [-sinθ, cosθ], which are the columns of A. Now we can do the multiplication Ar to finish. Ar gives you the image of r upon the transformation (in this case rotation specifically), which is what we seek.

Answer 5

The parametric equation for an ellipse with major axis 2a and minor axis 2b and center (0,0) is x = a cos t y = b sin t ..............t in R. If you tilt it such that the major axis makes an angle θ with the x-axis, you just have to add θ to t: x' = a cos(t+θ) y' = b sin(t+θ) ........t in R. Note that in this case, θ is constant. Edit: x' = a cos(t+θ) = a cos t cos θ - a sin t sin θ = x cos θ - a sin t sin θ y' = b sin(t+θ) = b sin t cos θ + b cos t sin θ = y cos θ + b cos t sin θ Well, let's say if t = 0, then x = a y = 0 and x' = a cos θ y' = b sin θ Do you see the difference? I agree that we can set a new parameter v = t+θ where θ is constant and get the exactly same parametric equation for the tilted ellipse, but I can't really find the flaw in my logic. Edit: I think I see my flaw in my logic... Consider any point (x,y) on the ellipse with center (0,0). x = p cos t y = p sin t where p is the distance between the point and the origin and t is the angle in between the segment with length p and the x-axis. If we rotate the coordinates with an angle of θ, we end up with x' = p cos(t+θ) = p cos t cos θ - p sin t sin θ = x cos θ - y sin θ = (a cos t) cos θ - (b sin t) sin θ y' = p sin(t+θ) = p sin t cos θ + p cos t sin θ = y cos θ + x sin θ = (b sin t) cos θ + (a cos t) sin θ but I still don't think this is the right one... it would be correct if it is a circle with r = b = a. I do see the rotated coordinates of, say, (a,0) is different from (a cos θ, b sin θ) since the rotation is circular while our figure is an ellipse. I will get back to you. Edit: I believe x' = p cos(t+θ) = p cos t cos θ - p sin t sin θ = x cos θ - y sin θ = (a cos t) cos θ - (b sin t) sin θ y' = p sin(t+θ) = p sin t cos θ + p cos t sin θ = y cos θ + x sin θ = (b sin t) cos θ + (a cos t) sin θ t in R is the correct solution.