# How do you identity if the equation 3x^2-2y^2+32y-134=0 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

##### 1 Answer
May 30, 2017

This is a hyperbola! noticing the negative ${y}^{2}$ term and the positive ${x}^{2}$ term.

#### Explanation:

This is a hyperbola in the general form, which means you must complete the square before graphing.

$3 {x}^{2} - 2 {y}^{2} + 32 y - 134 = 0$

$\therefore 3 {x}^{2} - 2 \left({y}^{2} - 16 y\right) = 134$

$\therefore 3 {x}^{2} - 2 \left({y}^{2} - 16 y + 64 - 64\right) = 134$

$\therefore 3 {x}^{2} - 2 {\left(y - 8\right)}^{2} + 128 = 134$

$\therefore 3 {x}^{2} - 2 {\left(y - 8\right)}^{2} = 6$

$\therefore \frac{{x}^{2}}{2} - \frac{{\left(y - 8\right)}^{2}}{3} = 1$

This is the standard form for a hyperbolic graph, because it is of the form:
$\frac{{\left(x - h\right)}^{2}}{a} ^ 2 - \frac{{\left(y - k\right)}^{2}}{b} ^ 2 = 1$

Of course, in this case $h = 0$, $k = 8$, $a = \sqrt{2}$ and $b = \sqrt{3}$.

The formula for the two oblique asymptotes are given by:

$y = \pm \frac{b}{a} \left(x - h\right) + k$

$\therefore y = \pm \frac{\sqrt{3}}{\sqrt{2}} \cdot x + 8$

$\therefore y = \frac{\sqrt{6} x}{2} + 8$ or $y = - \frac{\sqrt{6} x}{2} + 8$.

Now you can sketch the two asymptotes on the plane.

After this, we must find any x- and y-intercepts, simply by letting one or the other equal zero.

Letting $x = 0$, we get $- 2 {y}^{2} + 32 y - 134 = 0$, a quadratic which can be solved easily using the quadratic formula.

Letting $y = 0$, we get $3 {x}^{2} - 134 = 0$, another quadratic.

Now you are ready to sketch this hyperbola. Remember not to direct the curve away from the asymptote as it approaches it.