SOLUTION: Determine the center,foci,vertices,and covertices of the ellipse with the given equation: 9x^2+16y^2-126x+64y=71

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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