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]}