Question

The equation of a hyperbola is shown. What are the coordinates of the foci?

Answer 1

\(\bf \cfrac{(x-3)^2}{81}-\cfrac{y^2}{144}=1\implies \cfrac{(x-3)^2}{9^2}-\cfrac{(y-0)^2}{12^2}=1\)

so... where is the center of it?

Answer 2

(-15, 3) and (15, 3)

Answer 3

is that right? @jdoe0001

Answer 4

hmm the "center" is at the (h, k) of the equation , that is \(\bf \cfrac{(x-3)^2}{81}-\cfrac{y^2}{144}=1\implies \cfrac{(x-\color{red}{3})^2}{9^2}-\cfrac{(y-\color{red}{0})^2}{12^2}=1
\)

Answer 5

ohhh ok so what do I do now?

Answer 6

@jdoe0001

Answer 7

well, the foci are "c" distance from the center of the hyperbola

what's "c"? well \(\bf \cfrac{(x-h)^2}{a^2}-\cfrac{(y-k)^2}{b^2}=1\qquad \textit{"c" distance } = \sqrt{a^2+b^2}\\ \quad \\
c=\sqrt{9^2+12^2}\)

Answer 8

so the focci is -12? @jdoe0001

Answer 9

somewhat like so