Question

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?

Answer 1

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`