Donna,

There are both x² and y² terms so it's not a parabola. The fact that the coefficients of the x² and y² terms are both positive means it's not a hyperbola. Since the coefficients of both the x² and y² terms are the same (they are both 3) it's probably a circle. All that remains is to put it in standard form

(x - h)² + (y - h)² = r²

to check that the right side is positive. If the right side is zero then the radius of the circle is zero and the conic is just a point. If the right side is negative then there are no points on the graph of 3x² + 3y² - 4y - 8 = 0.

To put the equation in standard form first divide through by 3 to get

x² + y² - 4/3 y - 8/3 = 0

Now complete the square of the expression y² - 4/3 y - 8/3 = 0. This will allow you to write

x² + y² - 4/3 y - 8/3 = 0

in the form

(x - 0)² + (y - k)² = ?

If the right side is positive then you have a circle.

I hope this helps,
Harley