Question

How do you identity if the equation `y^2-2x^2-16=0` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

Please see below.

Explanation:

Let us write the equation in general form

`y^2-2x^2=16`

or `y^2/16-x^2/8=1`

or `y^2/4^2-x^2/(2sqrt2)^2=1`

It is in general form of hyperbola with center `(0,0)`. It is a vertical hyperbola as major axis is parallel to `y`-axis (`4` being greater than `2sqrt2`)

The general equation of such a vertical hyperbola is

`y^2/b^2-x^2/a^2=1`, then

vertices are `(0,+-b)`, focii are `(0,+-c)`, where `c^2=a^2+b^2`

and directrix are `x=+-b^2/c` and asymptotes are `y=+-b/ax`

Here vertices are `0,+-4)`, and as `c=sqrt(4^2+(2sqrt2)^2)=2sqrt6`, focii are `(0,+-2sqrt6)`.

Asymptotes are `y=+-sqrt2x`

The graph appears as follows:

graph{(y^2-2x^2-16)(y+sqrt2x)(y-sqrt2x)=0 [-40, 40, -20, 20]}