# Calculate the angle of an ellipse from it's matrix representation?

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 andrew gerzso 1 Vote Hello,I have generated a bunch of points then fed them to : ellipse =BoundingRegion[points, "FastEllipse"]. The output is : Ellipsoid[{0, 0}, {{0.0698865, 0.0545128}, {0.0545128, 0.0698865}}]. My question is: using the weighted matrix in the output – {{0.0698865, 0.0545128}, {0.0545128, 0.0698865}} – how can I calculate the angle of the ellipse ?Many Thanks, Andrew
12 days ago
5 Replies
 Michael Rogers 2 Votes Are these the angles you're after?: ArcTan @@@ Eigenvectors@{{0.0698865, 0.0545128}, {0.0545128, 0.0698865}} %/Degree (* {0.785398, 2.35619} {45., 135.} *) 
12 days ago
 Hello Michael, This is definitely a step forward for me ! However I am not sure how to interpret negative results. For example: ArcTan @@@ Eigenvectors@{{0.0749932, -0.128978}, {-0.128978, 0.406223}} %/Degree Out[93]= {1.90163, -2.81076} Out[94]= {108.955, -161.045} <<<<<<<<<<<<<<< here Many thanks ! Andrew
 Your answer goes in the right direction I think. But for example I am not sure how to interpret a negative angle such as in: ArcTan@@@Eigenvectors@{{0.0749932,-0.128978},{-0.128978,0.406223}} %/Degree Out[97]= {1.90163,-2.81076} Out[98]= {108.955,-161.045} <<< this negative angle. In any case thank you for your answer !
 Michael Rogers 2 Votes The range on ArcTan[x,y] per the docs is -Pi to Pi. Negative angles represent angles measured clockwise from the positive x axis, as in standard trigonometry. They will lie in quadrants III and IV if they are between -Pi and 0. You can add Pi or a 180 degrees to them and get a collinear angle that corresponds to the angle of the eigenvector -v, if v is the vector returned by Eigenvectors[] corresponding to the negative angle. (Any nonzero multiple of an eigenvector is also an eigenvector.)