How do you find the coordinates of the center for the given ellipse #4x^2+16x+352+16y^2+160y=0#?

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Answer 1 · 2015-12-23T17:03:16

#(-2,-5)#

Rearrange:

#(4x^2+16x)+(16y^2+160y)=-352#

Complete the square after factoring out the coefficient on the squared terms.

#color(red)4(x^2+4x+color(red)4)+color(blue)16(y^2+10y+color(blue)25)=-352+color(red)16+color(blue)400#

#4(x+2)^2+16(y+5)^2=64#

#(x+2)^2/16+(y+5)^2/4=1#

This is in the standard form of an ellipse:

#(x-h)^2/a+(y-k)^2/b=1#

Where the ellipse has a center at #(h,k)#.

The trick to finding the center is remembering to switch the sign of whatever's inside the parentheses:

#(x+2)^2rarrx"-coordinate " color(indigo)(-2)#
#(y+5)^2rarry"-coordinate "color(indigo)(-5)#

#(-2,-5)#

graph{(4x^2+16x)+(16y^2+160y)=-352 [-11.25, 8.75, -8.04, 1.96]}