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?

1 Answer
Jan 22, 2018

Answer:

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]}