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

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