Please write the equation of the conic section given the following information?: A hyperbola with vertices #(0,-6)# and #(0,6)# and asymptotes #y=3/4x# and #y=-3/4x#

1 Answer
Oct 28, 2017

#y^2/36 - x^2/64 = 1#

Explanation:

The vertices of the hyperbola are told to be at (0, -6) and (0,6). Since these have different y-coordinates, but the same x-coordinate, we know that the vertices are located above and below each other, making this a vertical hyperbola.

The standard form of a vertical hyperbola is given by:

#(y-y_c)^2/b^2 - (x-x_c)^2/a^2 = 1#

Center: #(x_c,y_c)#.

The center, by definition, is located in between the two vertices, or at (0,0) in this problem. This gives us:

#y^2/b^2 - x^2/a^2 = 1#

Furthermore, the equations of the asymptotes for the hyperbola is given by:

#(y-y_c) = +-b/a(x-x_c)#

Since we know that the asymptotes for this hyperbola are #y = +-3/4 x#, we know that the ratio #b/a = 3/4#. This doesn't mean that #b = 3# and #a = 4#, though; it just means that we could reduce #b/a# down to #3/4#.

If we keep in mind that the vertices are points on the graph, we know that (0,6) and (0,-6) both satisfy the equation. If we substitue (0,6) into our equation:

#6^2/b^2 - 0^2/a^2 = 1#

#36/b^2 = 1#

#36 = b^2 => b = 6#

Since #b/a = 3/4#, we know that #6/a = 3/4#, which leads us to #a = 8#. Our final equation:

#y^2/36 - x^2/64 = 1#

Graph:

graph{(y^2/36 - x^2/64 - 1)(3/4x-y)(-3/4x-y) = 0 [-20, 20, -10.5, 10.5]}