How do you find the standard form of $3x^2+5y^2-12x+80y+40=0$ ?

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Answer 1 · 2017-01-14T17:48:59

$frac{(x-2)^2}{5}+frac{(y+8)^2}{3}=24.8$

$3x^2+5y^2-12x+80y+40=0$

Combine x terms and y terms together:
$[3x^2-12x]+[5y^2+80y]+40=0$

Complete the square for x terms and y terms separately:
$3[x^2-4x]+5[y^2+16y]+40=0$

$3[(x-2)^2-4]+5[(y+8)^2-64]-40=0$

$3(x-2)^2-12+5(y+8)^2-320-40=0$

Combine constants and simplify:
$3(x-2)^2+5(y+8)^2-372=0$

$color(red)(1/15*)(3(x-2)^2+5(y+8)^2)=(372)color(red)(*1/15)$

$frac{3(x-2)^2}{15}+frac{5(y+8)^2}{15}=124/5$

$frac{(x-2)^2}{5}+frac{(y+8)^2}{3}=24.8$

How do you find the standard form of 3x^2+5y^2-12x+80y+40=03x^2+5y^2-12x+80y+40=0?