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

1 Answer
Feb 24, 2017

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

Explanation:

(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)