# 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?

Jan 22, 2018

#### Explanation:

Let us write the equation in general form

${y}^{2} - 2 {x}^{2} = 16$

or ${y}^{2} / 16 - {x}^{2} / 8 = 1$

or ${y}^{2} / {4}^{2} - {x}^{2} / {\left(2 \sqrt{2}\right)}^{2} = 1$

It is in general form of hyperbola with center $\left(0 , 0\right)$. It is a vertical hyperbola as major axis is parallel to $y$-axis ($4$ being greater than $2 \sqrt{2}$)

The general equation of such a vertical hyperbola is

${y}^{2} / {b}^{2} - {x}^{2} / {a}^{2} = 1$, then

vertices are $\left(0 , \pm b\right)$, focii are $\left(0 , \pm c\right)$, where ${c}^{2} = {a}^{2} + {b}^{2}$

and directrix are $x = \pm {b}^{2} / c$ and asymptotes are $y = \pm \frac{b}{a} x$

Here vertices are 0,+-4), and as $c = \sqrt{{4}^{2} + {\left(2 \sqrt{2}\right)}^{2}} = 2 \sqrt{6}$, focii are $\left(0 , \pm 2 \sqrt{6}\right)$.

Asymptotes are $y = \pm \sqrt{2} x$

The graph appears as follows:

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