Posted by Lucy on Sunday, October 12, 2008 at 7:57pm.
Find the coordinates of the center, the foci, the vertices, and the equations of the asymptotes of the graph of the equation: (x+1)^2/2y^2/8=1.
Center: (1,0)
foci:(1+/sqrt1, 0)
vertices: (1+/sqrt2, 0)
equations of the asymptotes: y0=sqrt8/sqrt2(x+1)
Is this correct and if not, how do I figure this out??

PreCalculus  Damon, Sunday, October 12, 2008 at 9:07pm
hyperbola due to  sign
yes, center at 1,0 because of(x+1)^2 and (y)^2
a =sqrt 2
b = sqrt 8 = 2 sqrt 2
so vertices at y =0 x = 1+sqrt 2 and 1 sqrt 2
center to focus = sqrt(2+8) = sqrt 10
so at y = 0 and x = 1 +/ sqrt 10
slope of asymptotes = +/ b/a = sqrt2
so
y = (+/ sqrt 2)  1
so yes, correct