Question

Identify the conic section that is represented by the equation:

x2 - 3y2 - 6x + 2y - 5 = 0.

a. hyperbola
b. Parabola
c. Ellipse
d. circle

Answer 1

helppppp..?

Answer 2

Based on that equation, you have a hyperbola. Notice that you could rearrange that to give the following:
\[x^2 -3y^2-6x + 2y -5 = 0\]
\[(x^2 - 6x) - 3(y^2 - \frac{2}{3}y) - 5 = 0\]
\[(x-3)^2 - 3(y-\frac{1}{3})^2 - 5 -9 - \frac{1}{3} =0 \]
\[(x-3)^2 - 3(y - \frac{1}{3})^2 - \frac{43}{3} = 0\]
If there was a plus sign between the x and y terms, you would have an ellipse. Since it's negative, you've got an hyperbola.

Answer 3

Oops, shoot, that last equation should have 41/3, not 43/3....other than that, it's fine.