Question 39159
Graph: 
16x^2+25y^2+64x-250y+289=0 

This is an ellipse. You have to convert the equation 
to a standard form by completing the square, as follows:

16(x^2+4x+4)+25(y^2-10y+25)=-289+16(4)+25(25)
16(x+2)^2 + 25(y-5)^2 = 400
Divide thru by 400 to get:
(x+2)^2/25 + (y-5)^2/16 = 1

Now it can be seen that the center of the ellipse 
is at (-2,5)
The major axis is 2(sqrt(25))=2(5)=10
The minor axis is 2(sqrt(16)=2(4)=8
So the vertices of the ellipse are at (-2-5,5)=(-7,5) and (-2+5,5)=(3,5)
With the information in the form you can also determine
the foci of the ellipse.
Hope this helps.
Cheers,
Stan H.