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?

1 Answer
Dec 27, 2016

Hyperbola (x-2)^2/6-(y+1)^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

ax^2+2hxy+by^2+ lower degree terms=0

represents an ellipse, parabola according as

ab-h^2>, = or < 0, respectively.

If a = b and h = 0, it represents a circle.

Here, ab-h^2=(6)(-5)-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

(x-2)^2/6-(y+1)^2/5=1.

Center: C( 2, -1)

Asymptotes: From
(x-2)^2/6-(y+1)^2/5=0

Separately,

(x-2)/sqrt6+-(y+1)/sqrt5=0