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 $\displaystyle 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.

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

Wouldn't i arrange like terms

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

then, completing the square has me baffled.

First term i take half of $\displaystyle -4 = 2^2 = 4$

so i now have

$\displaystyle (4x^2 -4x + 4) + (4y^2 - 3) = 0 + 4 $

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

If true then:

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

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

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

simplifying the left side further

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

Any help is much appreciated..

Thank you

Re: Classify the graph of the equation

Quote:

Originally Posted by

**icelated** Heres the equation.

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

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

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

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

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