Ellipse bounding a rectangle [closed]

Answer 1

Assuming you mean circumscribed (which is more precise than "enclosed"), you can read about how to circumscribe a rectangle here. From there, you can stretch it to rectangular, as Alnitak says.

Answer 2

1. Ellipse formula is (x/A)^2+(y/B)^2=1, where A and B are radiuses of ellipse
2. Rectangle sides are Rw and Rh
3. Let's assume we want ellipse with same proportions as rectangle; then, if we image square in circle (A=B,Rq=Rh) and squeeze it, we well keep ratio of ellipse A/B same as ratio of rectangle sides Rw/Rh;

This leads us to following system of equations:
(x/A)^2+(y/B)^2=1
A/B=Rw/Rh

Lets solve it: A=B*(Rw/Rh)
(Rh/2B)^2+(Rh/2B)^2=1
Rh=sqrt(2)*B

And final solution:
A=Rw/sqrt(2)
B=Rh/sqrt(2)

Example:

Answer 3

Experimentally, I found that an ellipse defined by a rectangle that is sqrt(2) larger than the inner rectangle works. So pass sqrt(2) to this function, and you will get the appropriate rectangle:

RectangleF boundingEllipse = GetScaledRectangle(innerRect, Convert.ToSingle(Math.Sqrt(2d)));

private RectangleF GetScaledRectangle(RectangleF rect, float scale) { float width = rect.Width * scale; float height = rect.Height * scale;

float gap = width - rect.Width;
float left = rect.Left - (gap / 2f);

gap = height - rect.Height;
float top = rect.Top - (gap / 2f);

return new RectangleF(left, top, width, height);

}

Answer 4

If you give your ellipse the same aspect ratio as the rectangle, you can work on the basis that what you want is a circle enclosing a square then stretched as if you've transformed the square into the required rectangle.

For a square with half side length = 1, the radius of the circle would be sqrt(2).

So, sweeping theta from 0 - 360', the ellipse's coordinate points will be:

• x = cos(theta) * sqrt(2) * rect.width + x.center;
• y = sin(theta) * sqrt(2) * rect.height + y.center;

where rect.width and rect.height are the half widths of the relevant sides.

Answer 5

The equation for a ellipse centered in the origin is

(x/A)^2 + (y/B)^2 = 1

Now if you want to enclose a rectangle of MxN with a eclipse you can move its center to the origin of coordinates. The top right coordinates are (M/2,N/2), replacing in the ellipse equation you have a formula you can use to solve B given A (or A given B).

If you have a rectangle of 4x2, the top-right coordinates are (2,1), replacing you have the (2/A)^2 + (1/B)^2 = 1, then if A=4 solving for B gives B=1/sqrt(1-(1/2)^2).