How do you find the center, vertices, foci and eccentricity of #x^2/100 + y^2/64 = 1#?

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Answer 1 · 2015-11-11T13:34:36

The ellipse is in the form

#(x - h)^2/a^2 + (y - k)^2/b^2 = 1#

Where

center #= (h, k)#

focus = #(h + c, k), (h - c, k)#

#c^2 = a^2 - b^2#

#x^2/100 + y^2/64 = 1#

#=> (x - 0)^2/10^2 + (y - 0)^2/8^2 = 1#

#=># center #= (0, 0)#

#c^2 = a^2 - b^2#

#c^2 = 100 - 64#

#=> c^2 = 36#
#=> c = 6#

#=># focus #= (h + c, k), (h - c, k)#

#=># focus #= (0 + 6, 0), (0 - 6, 0)#

#=># focus #= (6, 0), (-6, 0)#

I don't know how to get the eccentricity.
But I believe it has something to do with the ratio of #a# and #b# (or maybe #c#). Sorry