How do you find the vertices, asymptote, foci and graph #x^2/144-y^2/121=1#?

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Answer 1 · 2018-07-07T15:29:48

Please see the explanation below

This is the equation of a hyperbola

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

Here, we have

#x^2/12^2-y^2/11^2=1#

The center is #C=(h,k)=(0,0)# at the origin.

The vertices are

#V=(h+a,k)=(12,0)#

and

#V'=(h-a,k)=(-12,0)#

The equations of the asymptotes are

#y=k+-b/a(x-h)#

#y=+-11/12x#

That is

#y=-11/12x# and #y=11/12x#

Calculate

#c=sqrt(a^2+b^2)=sqrt(144+121)=sqrt(265)#

The foci are #F=(sqrt265, 0)# and #F'=(-sqrt265,0)#

And the graph is

graph{(x^2/144-y^2/121-1)(y-11/12x)(y+11/12x)=0 [-65.8, 65.84, -32.95, 32.9]}