# What type of conic section has the equation 9y^2 - x^2 - 4x+54y +68= 0?

Sep 20, 2014

9y^2−x^2−4x+54y+68=0 will have a hyperbola for its graph.

How do I know? Just a quick check of the coefficients on the ${x}^{2}$ and the ${y}^{2}$ terms will tell...
1) if the coefficients are both the same number and the same sign, the figure will be a circle.
2) if the coefficients are different numbers but the same sign, the figure will be an ellipse.
3) if the coefficients are of opposites signs, the graph will be a hyperbola.

Let's "solve" it: $- 1 \left({x}^{2} + 4 x\right) + 9 \left({y}^{2} + 6 y\right) = - 68$
Notice that I factored out the leading coefficients already, and gathered together the terms that both have the same variable.

$- 1 \left({x}^{2} + 4 x + 4\right) + 9 \left({y}^{2} + 6 y + 9\right) = - 68 + - 1 \left(4\right) + 9 \left(9\right)$
In this step, I completed the square by adding 4 and 9 inside of the parentheses, but then added to the other side, those numbers multiplied by the factored out numbers -1 and 9.

$- 1 {\left(x + 2\right)}^{2} + 9 {\left(y + 3\right)}^{2} = 9$ Rewrite in factored forms on the left.

$- 1 {\left(x + 2\right)}^{2} / 9 + {\left(y + 3\right)}^{2} / 1 = 1$ which just looks awkward...so I will change the order and make it look like subtraction:

${\left(y + 3\right)}^{2} - \frac{x + 2}{9} = 1$
That is what I wanted to see; I can tell what the center of the hyperbola is (-2,-3), how far to move from center to get to vertices (up and down 1 unit since the y-term is divided by 1) and the slope of the asymptotes ($\pm \frac{1}{3}$) . The "flatness" of this slope, in addition to the upward and downward opening of the curves, will make this graph fairly wide open.