I replied to that question yesterday.
Do you agree with my equation?
Since b > a , aren't the foci on the y-axis?
I had
b^2 = a^2 + c^2
9/4 = 1 + c^2
5/4 = c^2
c = ± √5/2
then e = c/b = (√5/2)/(3/2) = √5/3 or appr. .745
if e > 1 ,like you had, you would have a hyperbola
Find the center, vertices, foci, and eccentricity of the ellipse.
9x^2 + 4y^2 - 36x + 8y + 31 = 0
I know the center is (2,-1)
For the vertices I had (3,-1)(1,-1) and the foci (3.8,-1)(.20,-1) and e = 1.80 but I think these are wrong.
4 answers
Yesterdays post
http://www.jiskha.com/display.cgi?id=1272908984
http://www.jiskha.com/display.cgi?id=1272908984
so the vertices would be
(2, -1+sqrt5/2) (2, -1-sqrt5/2)?
(2, -1+sqrt5/2) (2, -1-sqrt5/2)?
yes