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

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Answer 1 · 2017-02-24T16:50:26

A single point is equivalent to that equation: #(1, 0)#

#(4x^2 - 8x) + y^2 = -4#

#(2x - a)^2 - a^2 + y^2 = -4# where #2*2x*a = 8x#

#a = frac{8x}{4x} = 2#

#(2x - 2)^2 + y^2 = -4 + 2^2#

#(x - 1)^2 + y^2/4 = 0#

#[sqrt 2 (x - 1)]^2 + y^2 = 0#

Whenever #y^2 + z^2 = 0#, #y = z = 0#

#x = 1# and #y = 0#

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If #y, z# where complex, then

#y^2 = -z^2#

#y = ±i z#

#z^2 + (± i z)^2 = 0#, for all #z in CC#

#forall x in CC, y(x) = ±i sqrt 2 (x - 1)#