Question

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola `(x+6)^2/36-(y+3)^2/9=1`?

Answer 1

Vertices are at `(-12,-3) and (0,-3)` Focii are at
`((-6-3 sqrt 5),-3) and ((-6+3 sqrt 5),-3)` and
asymptotes are `y= 1/2 x and y= -1/2 x -6`

Explanation:

`(x+6)^2/6^2- (y+3)^2/3^2=1; h=-6,k=-3,a=6 , b= 3`

This is standard form of the equation of a hyperbola with center

`(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :.` Center `(-6,-3)`

The vertices are `a` units from the center, and the foci are

`c` units from the center. Moreover `c^2=a^2+b^2` or

`c^2=6^2+3^2 :. c^2=45 or c = +- 3sqrt5`

Vertices are at `(-12,-3) and (0,-3)`

Focii are at `((-6-3 sqrt 5),-3) and ((-6+3 sqrt 5),-3)`

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions `2 a=12 and 2 b=6` with its

center at `(-6,-3) :.` slope `=+-b/a=+- 3/6= +- 1/2`.

Equation of asymptotes are `y+3= +-1/2(x+6)` or

`y=-3+-1/2(x+6) or y= 1/2 x and y= -1/2 x -6`

Therefore, asymptotes are `y= 1/2 x and y= -1/2 x -6`

graph{(x+6)^2/36-(y+3)^2/9=1 [-80, 80, -40, 40]}