SOLUTION: I was wonder how to solve the problem for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I was wonder how to solve the problem for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0).      Log On


   

Question 528160: I was wonder how to solve the problem for an ellipse if the endpoints of the major axis are at (1,6) and (1,-6) and the endpoints of the minor axis are at (5,0) and (-3,0).
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The center of your ellipse is obviously at (1,0), and the axes are vertical and horizontal, so it should be easy.
A circle centered at (1,0) and passing through (1,6) and (1,-6) would have a radius of 6, and the equation
%28x-1%29%5E2%2By%5E2=6%5E2 or
%28x-1%29%5E2%2F6%5E2%2By%5E2%2F6%5E2=1
A circle centered at (1,0) and passing through (-3,0) and (5,0) would have a radius of 4, and the equation
%28x-1%29%5E2%2By%5E2=4%5E2 or
%28x-1%29%5E2%2F4%5E2%2By%5E2%2F4%5E2=1
The ellipse centered at (1,0) and passing through (1,6), (1,-6), (-3,0) and (5,0) would have the equation
%28x-1%29%5E2%2F4%5E2%2By%5E2%2F6%5E2=1
In my mind it's just a deformed circle. To check that you got the right equation, substitute the cooordinates of the center one at a time. Substituting x=1 gives you the end points of the major axis. Substituting y=0, gives you the endpoints of the minor axis.
I do not remember what name the 4 and 6 are called, but they usually are represented with the letters a and b, and each one is half the length of one of the axes of the ellipse.