Question 1043332: Determine the center,foci,vertices,and covertices of the ellipse with the given equation:
9x^2+16y^2-126x+64y=71
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! 
Thge expressions in brackets remindme of some squares:
 , so  , and
 , so  .
Adding  to both sides of the equal sign in the original equation, we have
Dividing both sides of the equal sign in the equation above by  , we have
Dividing both sides of the equal sign in the equation above by  , we have
 , or  .
That is the equation of an ellipse centered at  ,
the point with  .
The semi-major axis length is  , and
semi-minor axis length is  .
The  is dividing the term with  ,
so the major axis is parallel to the x-axis,
and since the center has  ,
the major axis is the line .
On that major axis, are the vertices and foci.
The minor axis is parallel to the y-axis,
and since the center has  ,
the major axis is the line .
We know that the focal distance of an ellipse,  , can be found as
 , so in this case
 . An approximate value is  .
The locations of the foci are on the major axis,
at a distance  to either side of the center,
so one the coordinates of the foci are
 , or about  for one focus,
and  , or about  for the other focus.
Similarly, the vertices are on the major axis,
at a distance to either side of the center,
so one the coordinates of the vertices are
 ,
and  .
The covertices are on the minor axis,
at a distance to either side of the center,
so one the coordinates of the vertices are
 ,
and  .
The ellipse with axes and foci looks like this:
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