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Calculus III: Find the center, foci, vertices, and eccentricty of the ellipse, and sketch its graph. (x+4)^2+(y+6)^2 -------- = 1 1/4
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\[(x+4)^{2}+(y+6)^{2} = 1\] I don't know how to make a fraction using this code, but there is suppose to be a 1/4 underneath the (y+6)^2.
I don't know why its not posting up my equation in the proper form but anyways my problem with this particular problem is identifying "a" and "b". Would I have to flip the "1/4" first and then "a" and "b" would be 1?
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