# Classify the graph of the equation

• Dec 7th 2011, 01:03 PM
icelated
Classify the graph of the equation
I have an equation that i think is an ellipse or hyperbola since looking at the equation it has a $y^ and a x^2$.
However, i am having problems finishing it since i think i have to do complete the square but cant seem to get it.

Heres the equation.

$4x^2 + 4y^2 - 4x - 3 = 0$

Wouldn't i arrange like terms

$4x^2 -4x + 4y^2 - 3 = 0$

then, completing the square has me baffled.
First term i take half of $-4 = 2^2 = 4$

so i now have
$(4x^2 -4x + 4) + (4y^2 - 3) = 0 + 4$

would i then take half of 3 which is $(\frac {3}{2})^2 = \frac{9}{4}$

If true then:

$(4x^2 -4x + 4) + (4y^2 - 3 + \frac{9}{4}) = 0 + 4 + \frac{9}{4}$

pull out the 4 on the left side - but not ure how to handle the second side?

$4(x^2 -x + 1) + (4y^2 - 3 + \frac{9}{4}) = 0 + 4 + \frac{9}{4}$

simplifying the left side further

$4(x - 1)^2 + (4y^2 - 3 + \frac{9}{4}) = 0 + 4 + \frac{9}{4}$

Any help is much appreciated..
Thank you
• Dec 7th 2011, 01:27 PM
Plato
Re: Classify the graph of the equation
Quote:

Originally Posted by icelated
Heres the equation.
$4x^2 + 4y^2 - 4x - 3 = 0$

$4x^2 + 4y^2 - 4x - 3 = 0$

$x^2-x+y^2=\tfrac{3}{4}$

$x^2-x+\tfrac{1}{4}+y^2=\tfrac{3}{4}+\tfrac{1}{4}$

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