SOLUTION: I am not sure if you are able to show graphing, but if so I am having difficulty with this problem. Graph: 16x^2+25y^2+64x-250y+289=0 Thanks:)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I am not sure if you are able to show graphing, but if so I am having difficulty with this problem. Graph: 16x^2+25y^2+64x-250y+289=0 Thanks:)      Log On


   

Question 39159: I am not sure if you are able to show graphing, but if so I am having difficulty with this problem.
Graph:
16x^2+25y^2+64x-250y+289=0
Thanks:)
Found 2 solutions by Nate, stanbon:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
16x%5E2%2B25y%5E2%2B64x-250y%2B289=0
16x%5E2%2B64x%2B25y%5E2-250y=-289
16%28x%5E2%2B4x%29%2B25%28y%5E2-10y%29=-289
16%28x%2B2%29%5E2%2B25%28y-5%29%5E2=400
%2816%28x%2B2%29%5E2%2F400%29%2B%2825%28y-5%29%5E2%2F400%29=400%2F400
%28%28x%2B2%29%5E2%2F25%29%2B%28%28y-5%29%5E2%2F16%29=1
This is an ellipse with a center at (-2,5). It has a horizontal major axis. The major axis is ten units long while the minor axis is eight units long. I am sorry .... i do not know how to graph this via programs available to this web site.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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.