document.write( "Question 39159: I am not sure if you are able to show graphing, but if so I am having difficulty with this problem.
\n" ); document.write( "Graph:
\n" ); document.write( "16x^2+25y^2+64x-250y+289=0\r
\n" ); document.write( "\n" ); document.write( "Thanks:) \n" ); document.write( "
Algebra.Com's Answer #24615 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Graph:
\n" ); document.write( "16x^2+25y^2+64x-250y+289=0 \r
\n" ); document.write( "\n" ); document.write( "This is an ellipse. You have to convert the equation
\n" ); document.write( "to a standard form by completing the square, as follows:\r
\n" ); document.write( "\n" ); document.write( "16(x^2+4x+4)+25(y^2-10y+25)=-289+16(4)+25(25)
\n" ); document.write( "16(x+2)^2 + 25(y-5)^2 = 400
\n" ); document.write( "Divide thru by 400 to get:
\n" ); document.write( "(x+2)^2/25 + (y-5)^2/16 = 1\r
\n" ); document.write( "\n" ); document.write( "Now it can be seen that the center of the ellipse
\n" ); document.write( "is at (-2,5)
\n" ); document.write( "The major axis is 2(sqrt(25))=2(5)=10
\n" ); document.write( "The minor axis is 2(sqrt(16)=2(4)=8
\n" ); document.write( "So the vertices of the ellipse are at (-2-5,5)=(-7,5) and (-2+5,5)=(3,5)
\n" ); document.write( "With the information in the form you can also determine
\n" ); document.write( "the foci of the ellipse.
\n" ); document.write( "Hope this helps.
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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