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

1 Answer
Feb 8, 2016

Answer:

#4y^2-8y-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
#color(white)("XXX")4y^2-8y-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 #(x^2)# instead of simply #x# then the equation would be a hyperbola (and not an ellipse) so I'm not certain what was intended.