How do you find all the critical points to graph #-4x^2 + 9y^2 - 36 = 0# including vertices, foci and asymptotes?

1 Answer
Feb 16, 2018

Answer:

The center is located at the origin, vertices at #(0,2)# and #(0,-2)#, and foci at #(0,3.6)# and #(0,-3.6)#; the slopes of the asymptotes are #y^2=+-2/3x^2#
#Graph:# graph{y^2/4-x^2/9=1 [-10, 10, -5, 5]}

Explanation:

We can arrange this equation into #9y^2-4x^2=36# in order to correct format. I also added #36# to both sides as this will become necessary down the road. Next, I will divide both sides by #36#: #(9y^2)/36-(4x^2)/36=1#.
Simplify the remaining equation; #y^2/4-x^2/9=1#

The next step is to find #a#, #b#, and #c# (excluding the center; it is at the origin). In order to do so, we have to define where they are:
#y^2/a^2-x^2/b^2=1#

#a# is always the root of the first denominator in hyperbolic equations, so it would be #2#.

#b# would be #3# (#sqrt9=3#)

#c# can be found, only in hyperbolic equations, through Pythagora's Theorem: #a^2+b^2=c^2#
#c=~3.6#

To find the foci, add and subtract the value of #c# to the variable being divided by #a# (which is #y#). This value must be directly above or to the side of the center.
The value of the foci are #(0,3.6)# and #(0,-3.6)#

To find the vertices, add and subtract your #a# value to the number being divided by it (#y#, again).
Vertices: #(0,2)# and #(0,-2)#.

To find the slope depends on the format: if #y^2/a^2-x^2/b^2#, the slopes of the asymptotes are #+-a/b#. (If your equation looks like this: #x^2/a^2-y^2/b^2=1#, the slope will be #+-b/a#)
The slopes of the asymptotes are #+-2/3#