How do I graph #9x^2+16y^2-36x+32y=92# algebraically?

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Answer 1 · 2014-10-19T21:43:58

By grouping #x#'s and #y#'s together,

#Rightarrow (9x^2-36x)+(16y^2+32y)=92#

by factoring out 9 and 16,

#Rightarrow9(x^2-4x)+16(y^2+2y)=92#

by adding and subtracting 4 and 1 in the parentheses,

#Rightarrow9(x^2-4x+4-4)+16(y^2+2y+1-1)=92#

by keeping the first three terms in each group,

#Rightarrow9(x^2-4x+4)-36+16(y^2+2y+1)-16=92#

by completing the squares and adding 36 and 16,

#Rightarrow9(x-2)^2+16(y+1)^2=144#

by dividing by 144,

#Rightarrow(x-2)^2/{16}+(y+1)^2/{9}=1#

Hence, we have an equation of an ellipse

#(x-2)^2/{4^2}+(y+1)^2/{3^2}=1#,

which is 8 units wide, 6 units tall, and centered at #(2,-1)#.

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I hope that this was helpful.