Question

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`

Answer 1

`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]}