Question

How do you graph `(y+1)^2/36 - (x+5)^2/9 =1`?

Answer 1

`"center " = (-5, -1); " "a = 6; " "b = 3`
`"vertices "(-5, 5), (-5, -7)`

Explanation:

Given: `(y+1)^2/36 - (x+5)^2/9 = 1`

Form is a Vertical Transverse axis hyperbola: `(y-k)^2/a^2 - (x-h)^2/b^2 = 1`

`"center " = (-5, -1)`

`a = sqrt(36) = 6; " " b = sqrt(9) = 3`

Since the larger number in the denominator is over the `y` value, this is the `a`. Create a dashed box that is `a = 6` above and below the center in the `y` direction. Make it be `b = 3` to the right and left of the center.

Create dashed lines that connect the corners (asymptotes) and extend past the corners. The equations of the asymptotes are: `y = +- a/b(x-h)+k` for the vertical transverse hyperbola.

Let the vertices be at `(h, k+-6) = (-5, 5), (-5, -7)`

The hyperbola has two branches. Each branch goes through one vertex and stays inside the asymptotes.