How do you find the foci and sketch the hyperbola #x^2/9-y^2/4=1#?

We know that the standard Cartesian form for the equation of a hyperbola with a transverse horizontal axis,

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

has foci at #(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#

If we write the given equation in the same form as equation [1], then it is a simple matter to find the foci:

#(x - 0)^2/3^2- (y-0)^2/2^2 = 1" [2]"#

Now that we have the given equation in the same form, the computation for the foci is trivial:

#(h-sqrt(a^2+b^2),k)# and #(h+sqrt(a^2+b^2),k)#

#(0-sqrt(9+4),0)# and #(0+sqrt(9+4),0)#

#(-sqrt(13),0)# and #(sqrt(13),0) larr# these are the foci.

To graph the equation you will need the vertices:

#(h-a,k)# and #(h+a,k)#

#(0-3,0)# and #(0+3,0)#

#(-3,0)# and #(3,0) larr# these are the vertices.

And you will need the equations of the asymptotes:

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

#y = -2/3(x-0) + 0# and #y = 2/3(x-0) + 0#

#y = -2/3x# and #y = 2/3x larr# there are the equations of the asymptotes.

Here is a graph of, the equation, the foci, the vertices, and the asymptotes.