Question

Last hyperbola question (I hope) :)

Answer 1

you think you can help me

Answer 2

you seem smart

Answer 3

Finding the equation

42. Center at (-3, 1); focus at (-3, 6); vertex at (-3, 4)

I know I need to use \(\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2} = 1\)

6 - 1 = 5, c = 5
4 - 1 = 3, a = 3
\(b^2\) = 25 - 9
\(b^2\) = 16

\(\dfrac{(y-1)^2}{9}-\dfrac{(x+3)^2}{16} = 1\)

Answer 4

yup

Answer 5

Ah thin you has it.

Answer 6

Thank you both :)