Question

Find the vertices and foci of the hyperbola with equation
@agent0smith please

Answer 1

(x+5)^2/36-(y+1)/64=1

Answer 2

Find the center first: (-5, -1)

vertices \[\Large (-5 \pm a, -1)\]where a^2 is the number under the x^2 term. b^2 is under the y term.

foci \[\Large (-5 \pm c, -1)\] where c^2 = a^2+b^2

Answer 3

Vertices: (-1, 1), (-1, -11); Foci: (-1, -15), (-1, 5)?

Answer 4

Is the (y+1) term squared? (y+1)^2 ? Otherwise this is a parabola :)

Answer 5

yes @ranga

Answer 6

You seem to have mixed the x and y's up... see my post again.

Answer 7

@agent0smith so would it be:

Answer 8

Vertices: (1, -1), (-11, -1); Foci: (-15, -1), (5, -1)

Answer 9

Looks fine.