# How do you identity if the equation 6x^2-24x-5y^2-10y-11=0 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Dec 27, 2016

Hyperbola ${\left(x - 2\right)}^{2} / 6 - {\left(y + 1\right)}^{2} / 5 = 1$. Socratic asymptotes-inclusive graph is inserted graph{((x-2)^2/6-(y+1)^2/5-1)((x-2)^2/6-(y+1)^2/5)=0 [-10, 10, -5, 5]}

#### Explanation:

The second degree equation

$a {x}^{2} + 2 h x y + b {y}^{2}$+ lower degree terms=0

represents an ellipse, parabola according as

$a b - {h}^{2} > , = \mathmr{and} < 0$, respectively.

If $a = b \mathmr{and} h = 0$, it represents a circle.

Here, $a b - {h}^{2} = \left(6\right) \left(- 5\right) - 0 = - 30 < 0$, and so, the graph is a hyperbola.

As h = 0, the axes are parallel to the axes of coordinates.

So, the form is reduced to the standard form as

${\left(x - 2\right)}^{2} / 6 - {\left(y + 1\right)}^{2} / 5 = 1$.

Center: $C \left(2 , - 1\right)$

Asymptotes: From
${\left(x - 2\right)}^{2} / 6 - {\left(y + 1\right)}^{2} / 5 = 0$

Separately,

$\frac{x - 2}{\sqrt{6}} \pm \frac{y + 1}{\sqrt{5}} = 0$