How do you find the standard form given #4x^2+9y^2-8x+54y+49=0#?

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Answer 1 · 2017-02-24T08:27:47

#(2 x - 2)^2 + (3 y + 9)^2 = 36#

#4 x^2 + 9 y^2 - 8 x + 54 y + 49 = 0#

#(2 x)^2 + (3 y^2) - 8 x + 54 y + 49 = 0#

rearrange the equation,
#[(2 x)^2 - 8 x] + [(3 y^2) + 54 y] + 49 = 0#

#[(2 x - 2)^2 - 2^2] + [(3 y + 9^2) -9^2] + 49 = 0#

#(2 x - 2)^2 + (3 y + 9^2) -4 -81 + 49 = 0#

#(2 x - 2)^2 + (3 y + 9)^2 - 36 = 0#

#(2 x - 2)^2 + (3 y + 9)^2 = 36#