How do you graph #y^2-9x^2=9# and identify the foci and asympototes?

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Answer 1 · 2016-11-22T09:56:30

The foci are F#=(0,sqrt10)# and F'#(0,-sqrt10)#
The equations of the asymptotes are #y=3x# and #y=-3x#

Let's rewrite the equation as

#y^2/9-x^2/1=1#

This is the equation of an up-down hyperbola.

The general equation is

#(y-k)^2/a^2-(x-h)^2/b^2=1#

The center is #(h,k)=(0,0)#

The vertices are #(h,k+-a)#, A#=(0,3)# and A'#=(0,-3)#

The slope of the asymptotes are #+-a/b# #=+-3#

The equations of the asymptotes are #y=3x# and #y=-3x#

To determine the foci, we need c, #c^2=a^2+b^2#

So, #c=+-sqrt(a^2+b^2)=+- sqrt(9+1)=+-sqrt10#

And the foci are F#=(0,sqrt10)# and F'#(0,-sqrt10)#

graph{(y^2/9-x^2-1)(y-3x)(y+3x)=0 [-12.66, 12.65, -6.32, 6.34]}