Finding 8 points on an ellipse

Question

Tom 2021-9-20

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1456799-finding-8-points-on-an-ellipse

评论： Image Analyst 2021-9-20

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Dear experts,

I have created a 95% confidence ellipse using the following code around a dataset:

allX = score(:,1);
allY = score(:,2);
p = 95;
CIx = prctile(allX, [(100-p)/2, p+(100-p)/2]);
CIy = prctile(allY, [(100-p)/2, p+(100-p)/2]);
CIrng(1) = CIx(2)-CIx(1);
CIrng(2) = CIy(2)-CIy(1);
llc = [CIx(1),CIy(1)];
rectangle('Position',[llc,CIrng],'Curvature',[1,1], 'EdgeColor',"red");

Now I need to find 8 specific points on the ellipse. Points W - Z have to be the lowest, highest, most right and most left of the ellipse (as you can see here (those are normally not the problem, i think you don´t have to help me with these four points):

The points A-D are more of a problem for me. How can I find these? They are somehow the corners of the rectangle so the corners of the ellipse, but how do I find the coordinates of those?

Thanks in advance.

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Tom 2021-9-20

@Walter Roberson

Yes the area is equal. Sorry for not making it clear in advance.

Image Analyst 2021-9-20

Can you attach your data with the paperclip icon?

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Answer 1

Walter Roberson 2021-9-20

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1456799-finding-8-points-on-an-ellipse#answer_791159

Refering to https://math.stackexchange.com/questions/493104/evaluating-int-ab-frac12-r2-mathrm-d-theta-to-find-the-area-of-an-ellips

When the ellipse is parameterized as

then the area for the angle from 0 to θ is

And the total area of an ellipse of this form is

. We want a sector that is half of the upper quadrant. The upper quadrant by itself would have an area of

. So we want

. Factoring out the

and a 1/2 from both sides then we have

and so

. So

and then

. But tan pi/4 is 1, so

Checking with (say) a = 5, b = 2, then

. Compute 1/2 * a * b * atan(5/2*tan(atan(2/5))) = 1/2 * 5 * 2 * atan(5/2 * 2/5) = 5 * atan(1) = 5 * pi/4 . And on the left side, 1/2 * pi/4 * 5 * 2 = 5 * pi/4 .

So the formula for theta turns out to be very simple: atan2(b,a)

Now you just have to transform CIrng into a, b parameters.