Question

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

Answer 1

have you drawn the coordinates yet?

is the hyperbola of vertical or horizontal traverse axis?

Answer 2

vertical?

Answer 3

hmm yes, the "vertices" are at (3, -1) and (3, -9) so

Answer 4

any ideas where the center might be?

Answer 5

(3,-5)?

Answer 6

yes
so our center is at (3,-5)

our vertices are at (3, -1) and (3, -9)
the distance between (3, -1) and (3, -9) is well, \(\bf 8\)

our co-vertices are at (-6. -5) and (12, -5)
the distance between (-6. -5) and (12, -5) is, well \(\bf 18\)

the traverse axis will be 8, and component "a" is half that, or 8/2 = 4 = a
the conjugate axis is 18 and component "b" is half that or 18/2 = 9 = b

the hyperbola is vertical, so the fraction with the "y" variable wil be the positive one

\(\bf \cfrac{(y-k)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1
\\ \quad \\ \quad \\
a=4\qquad b=9\qquad center\ (3,-5)\to \cfrac{(y-(-5))^2}{4^2}-\cfrac{(x-3)^2}{9^2}=1
\\ \quad \\
\implies \cfrac{(y+5)^2}{4^2}-\cfrac{(x-3)^2}{9^2}=1\)