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

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Answer 1 · 2016-10-22T21:09:40

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

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]}