Question

How do you identify the vertices, foci, and direction of `(y+8)^2/36-(x+2)^2/25=1`?

Answer 1

Because the x term is negative, we know that the hyperbola opens vertically; the type of hyperbola is called a vertical transverse axis type; its standard form is:

`(y - k)^2/a^2 - (x - h)/b^2 = 1" [1]"`

The center of this type is the point `(h, k)`
The vertices are the points `(h, k - a) and (h, k + a)`
The foci of this are are the points `(h, k - sqrt(a^2 + b^2)) and (h, k + sqrt(a^2 + b^2))`

Rewrite the given equation in the form of equation [1]:

`(y - -8)^2/6^2 - (x - -2)/5^2 = 1" [2]"`

he center of this type is the point `(-8, -2)`
The vertices are the points `(-8, -8) and (-8, 4)`
The foci of this are are the points `(-8, -2 - sqrt(61)) and (-8, -2 + sqrt(61))`