How do you graph #12y^2-25x^2=300# and identify the foci and asympototes?

1 Answer
Jan 4, 2018

Answer:

The foci are #(0,3sqrt3)# and #(0,-3sqrt3)#
The asymptotes are #y=5/(2sqrt3)(x)# and #y=-5/(2sqrt3)(x)#

Explanation:

First, we see turn the equation so that it equals one.
#12y^2-25x^2=300 => (12y^2)/300-(25x^2)/300=1#
You can simplify this to:
# (y^2)/25-(x^2)/12=1#

Remember that a hyperbolic equation can be in two forms:
#x^2/a^2-y^2/b^2=1# or #y^2/a^2-x^2/b^2# and #c=sqrt (a^2+b^2)#

Now, a graph of a hyperbola opens to the left and the right when #x^2# is over #a^2#.
Also, a graph of a hyperbola opens to up and down when #y^2# is over #a^2#.

Because of this, the foci of a hyperbola is #(h+-c,k)# when #x^2# is over #a^2#.

Similarly, the foci of a hyperbola is #(h,k+-c)# when #y^2# is over #a^2#.

The equation for the asymptote of a hyperbola is #k+-b/a(x-h)# when #x^2# is over #a^2#.

Similarly, the equation for the asymptote of a hyperbola is #k+-a/b(x-h)# when #y^2# is over #a^2#

Since #y^2# is over #a^2#, we use our appropriate rules to get:
#c=sqrt (12+25) => c=3sqrt3#

Therefore, the foci are #(0,3sqrt3)# and #(0,-3sqrt3)#
The asymptotes are #y=5/(2sqrt3)(x)# and #y=-5/(2sqrt3)(x)#

I have a graph from desmos.com here to visualize it.
desmos.com