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## How to determine whether a points lies in an ellipse Tag : r , By : fayoh Date : November 24 2020, 05:47 AM Boards Message : You Must Login Or Sign Up to Add Your Comments .

## Determining if a point lies within an ellipse, including the edge

Tag : java , By : Lee KW
Date : March 29 2020, 07:55 AM
may help you . You have to decide what kind of tolerance your method will use. While your example uses points that are expressible in floating point, there are many points along the border of the ellipse which will not be, and so deciding whether a point is "on the border" isn't clear-cut. If you don't much care, then I would suggest making the ellipse slightly "bigger" than you actually want and using the built-in contains() method.
If you want to write your own method, it's as simple as taking the formula for an ellipse, plugging in the X and Y values of the point you wish to test, and observing the result:
``````bool isInsideOfOrOnBorderOfEllipse = ((x*x)/(a*a) + (y*y)/(b*b)) <= 1;
``````

## how to determine the rotation angle of ellipse and fill area between points in opencv, c++

Tag : cpp , By : hlpimfalling
Date : March 29 2020, 07:55 AM
Any of those help After short check - it seems that OpenCV function calculates middle angle as ma = (angle1 + angle2) / 2 and draws arc through this point.
Both (-45,45) and (45,-45) give the same 90-degrees arc through zero, both (315,45) and (45,315) give the same 270-degrees arc.

## Points of Intersection of an Ellipse and a line after rotating ellipse by angle theta

Tag : python , By : Novi Indrayani
Date : March 29 2020, 07:55 AM
it fixes the issue If you have tested and ready-to-use solution for axis-aligned ellipse, it is much simpler to transform points defining line into ellipse system, find intersection points, then make reverse transformation.
For origin-centered ellipse and rotation angle theta
`````` x1' = x1 * Cos(theta) + y1 * Sin(theta)
y1' = - x1 * Sin(theta) + y1 * Cos(theta)
``````
`````` x1' = (x1 - cx) * Cos(theta) + (y1 - cy) * Sin(theta)
y1' = - (x1 - cx) * Sin(theta) + (y1 - cy) * Cos(theta)
``````

## Given the mid-points of ellipse, its orientation, minor and major axis length, get n points on the ellipse

Tag : python , By : vferman
Date : March 29 2020, 07:55 AM
this one helps. You can use OpenCV to draw an ellipse.
``````cv2.ellipse(img,(256,256),(100,50),0,0,360,255,-1)
``````
``````gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
contours, _ = cv2.findContours(gray,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
epsilon = 0.01*cv2.arcLength(contours,True)
green_dots = cv2.approxPolyDP(contours,epsilon,True)
``````

## Can I determine, using only a subset of polygon points, where the interior of the polygon lies?

Tag : algorithm , By : Si Gardner
Date : March 29 2020, 07:55 AM
should help you out The usual approach would be to project a line horizontally left from your given point and find the nearest edge that that line intersects. The direction of that edge tells you whether the point is inside or outside the polygon (with a path direction imposed as per Paul's suggestion). If no edge is encountered, you are, of course, outside.
To achieve this, you need each cell of your binary tree to reference every edge that passes through the cell, even if its vertices are outside the cell. You may sometimes need to traverse through multiple cells to find the nearest intersecting edge. 