Question 603952: How do you write an ellipse in standard form when you only know the foci and the covertices?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Knowing the foci and the covertices is just enough.
The center of the ellipse is the midpoint of the line segment that connects the foci.
That should also be the midpoint of the segment that connects the covertices.
The focal distance, c, is half of the distance between the foci.
The semi-minor axis, b, is half the distance between the covertices.
The semi-major axis, a, is half the distance between the vertices, and can be calculated from b and c form the formula
If the covertices have the same x-coordinate, the minor axis that connects them is vertical and the equation of the ellipse will be
 with (h,k) being the center of the ellipse.
If the covertices have the same y-coordinate, then the minor axis is horizontal, and the equation would be
 .
(If the covertices do not have a coordinate in common, the equation is more complicated, and this is not high school math any more).
HOW I KNOW THAT :
In the diagram below, O is the center of the ellipse, with axes XA and BY.
(I did not draw the whole ellipse, just the important points).
 A and X are vertices; B and Y covertices; C and Z foci.
The important segment lengths are:
OA=OX=a (the semi-major axis)
OB+OY=b (the semi-minor axis)
OC=OZ=c (the focal distance
Covertex B is one of the points of the ellipse.
The distances from B to focus C and to focus Z are the same BZ=BC.
Their sum BZ+BC=2BC is the same as the sum of distances to the foci for all points on the ellipse.
Vertex A is one of the points on the ellipse.
The sum of its distances to the foci is AC+AZ=AC+(AO+OZ)=(a-c)+(a+c)=2a
2BC=2a --> BC=a
In the right triangle OBC, Pythagoras theorem says that
|
|
|
