# How do you find the standard form of 4y^2-8y-x+1=0?

Feb 8, 2016

$4 {y}^{2} - 8 y - x - 1 = 0$ is already in standard form...
but since this was asked under "Geometry of an Ellipse" the question was probably meant to be
$\textcolor{w h i t e}{\text{XXX}} 4 {y}^{2} - 8 y - {x}^{2} - 1 = 0$

#### Explanation:

A general polynomial is said to be in "standard form" if every term has a degree greater than or equal to all terms that follow it.

{:(4y^2,"degree "2),(-8y,"degree "1),(-x,"degree "1),(+1,"degree "0):}rArr "in standard form"

However certain "special polynomials" such as those representing ellipse or hyperbola have their own "standard forms".

If the given equation was meant to have $\left({x}^{2}\right)$ instead of simply $x$ then the equation would be a hyperbola (and not an ellipse) so I'm not certain what was intended.