Question

How do you find the vertices, asymptote, foci and graph `y^2/25-x^2/16=16`?

Answer 1

Vertices: `(0, +-20)`
Asymptotes: `y = +- 5/4 x`
Foci: `(0, +-4sqrt(41))`

Explanation:

First put the hyperbola in standard form when the largest denominator is under the `y^2`: `(y-k)^2/a^2 - (x-h)^2/b^2 = 1`

Multiple all terms by `1/16` to get a `1` on the right-side:
`y^2/(25*16) - x^2/(16*16) = 1`; `y^2/(400) - x^2/16^2 = 1`; `y^2/20^2 - x^2/16^2 = 1`

Center: `(h,k) = (0,0)`

`a = 20, b = 16`

`c^2 = a^2+b^2 = 656, c = sqrt(656) = sqrt(16*41) = 4sqrt(41)`

Vertices: `(h, k+-a) = (0, +-20)`

Asymptotes: `y-k = +-a/b(x-h); y = +-20/16x; y = +-5/4x`

Foci: `(h,k+-c) = (0, +-4sqrt(41))`

To graph:

1. Create a box centered around the center that is length `a` vertically above and below the center and length `b` horizontally above and below the center (use dashed lines).
2. Sketch the asymptote lines (use a dashed line).
3. Place dots at the vertices and foci.
4. Sketch a curve from each vertex, staying inside of the asymptote lines.
   Graphing Hyperbolas