Question

Did I do this right? Finding the equation of a hyperbola, center not at (0, 0)
Question below

Answer 1

41. Center at (4, -1); focus at (7, -1); vertex at (6, -1)

From that I figured out that I needed to use \(\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1\)

Since 7-4 is 3, c = 3
6-4 is 2
a = 2
\(b^2\) = 9-4
\(b^2\) = 5

\(\dfrac{(x-4)^2}4 - \dfrac{(y+1)^2}9 = 1\)

Answer 2

Whoops, that should be \(\dfrac{(y+1)^2}5\)

Answer 3

yes, it looks correct

Answer 4

Yay! Thank you :)

Answer 5

Well done!
http://prntscr.com/9ol6fa