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

Feb 24, 2017

A single point is equivalent to that equation: $\left(1 , 0\right)$

#### Explanation:

$\left(4 {x}^{2} - 8 x\right) + {y}^{2} = - 4$

${\left(2 x - a\right)}^{2} - {a}^{2} + {y}^{2} = - 4$ where $2 \cdot 2 x \cdot a = 8 x$

$a = \frac{8 x}{4 x} = 2$

${\left(2 x - 2\right)}^{2} + {y}^{2} = - 4 + {2}^{2}$

${\left(x - 1\right)}^{2} + {y}^{2} / 4 = 0$

${\left[\sqrt{2} \left(x - 1\right)\right]}^{2} + {y}^{2} = 0$

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 $z \in \mathbb{C}$

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