How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola #x^2-36y^2=36#?

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Answer 1 · 2017-12-21T15:08:29

Vertices are #(6,0)# and #(-6,0)#, focii are #(sqrt37,0)# and #(-sqrt37,0)# and asymptotes are #x+-6y=0#.

If the general equation of hyperbola is

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

vertices are #(h+-a,k)#, focii are #(h+-c,k)#, where #c^2=a^2+b^2#, directrix are #x=+-a^2/c#

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

Here equation is #x^2-36y^2=36# or #x^2/6^2-y^2/1^2=1#

as #a=6# and #b=1#, vertices are #(6,0)# and #(-6,0)#,

and #c=sqrt(36+1)=sqrt37#,

focii are #(sqrt37,0)# and #(-sqrt37,0)#

directrix are #x=+-36/sqrt37#

Asymptotes are #y=+-x/6# or #x+-6y=0#

graph{(x^2-36y^2-36)(x+6y)(x-6y)=0 [-10, 10, -5, 5]}