How do you find the standard form of #x^2+4y^2+10x+24y+45=0#?

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Answer 1 · 2016-11-29T12:04:22

#(x+5)^2 /4^2 +(y+3)^2 /2^2 =1#

Re write it as #x^2 +10x +4y^2 +24y+45=0#

#x^2 +10x +25 -25 +4(y^2 + 6y +9-9) +45=0#

#(x+5)^2 -25 +4(y^2 +6y+9) -36 +45=0#

#(x+5)^2 +4 (y+3)^2 -16=0#

#(x+5)^2 + 4(y+3)^2 = 16#

#(x+5)^2 / 16 + (y+3)^2 /4=1#

#(x+5)^2 /4^2 +(y+3)^2 /2^2 =1#

This represents en ellipse centered at (-5,-3) with semi-major axis 4 and semi -minor axis 2