How do you graph #y^2/16-x^2/4=1# and identify the foci and asympototes?

1 Answer
Oct 22, 2016

Answer:

The foci are #(0,+2sqrt5)# and #(0,-2sqrt5)#
And the asymptotes are #y=2x# and #y=-2x#

Explanation:

The graph is a hyperbola "opens up and down"

The center is #(0,0)#

The vertices are #(0,4)# and #(0,-4)#

The slopes of the asymptotes are #2# and#-2#

The equations of the asymptotes are #y=2x# and #y=-2x#

To determine the foci, we need #c=+-sqrt(16+4)=+-sqrt20=+-2sqrt5#

The foci are #(0,+2sqrt5)# and #(0,-2sqrt5)#

graph{(y/4)^2-(x/2)^2=1 [-40, 40, -20, 20]}