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

Answer:

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#