Question

Equation of a hyperbola with vertices (3, -1) and (3, -9) and co-vertices (-6. -5) and (12, -5).

Answer 1

you have the vertices, where is the center?

Answer 2

The center is not given.

Answer 3

but it lies in a particular place given the vertices.

Answer 4

Where the vertices meet, right?

Answer 5

do you mean between the vertices?

ok: it'll be at (3,-5).

Answer 6

Oh, okay. So I'd plug in the center point into the equation to get,
(x-3)^2 / a^2 - (y+5)^2 /b^2

Answer 7

ok, but the hyperbola is again vertical, so it's a y^2-x^2 = 1 type equation.
the number that divides (y+5)^2 is the distance to the vertex. (4)
the number that divides (x-3)^2 is the distance to the covertex. (9)

Answer 8

it is easy to remember who is a or b, who does what. the fraction (y-p)^2/q^2 uses only "y-related" numbers. So, the y-distance, the y-coordinate of the center.
Same for the (x-l)^2/m^2. (x-distance, x-coordinate).

Answer 9

yes, the center will be half-way in between vertices, as reemii showed above
so the center will be at (3, -5) as he pointed out

the "a" element will be the distance of the major traverse axis
it goes from 4 units up and 4 units down from (3, -5), so
a= 4

the "b" element will be the distance of the conjugate axis
you're given the so-called co-vertices, or the end-points for the conjugate axis
so it goes from (-6, -5) to (12, -5) moving over the y = -5 line
so -6 to 12, there are 18 units, so it moves to the left 9 units and 9 to the right
so
b = 9