Posted by confused on Monday, October 5, 2009 at 9:01pm.
first find the centre, which would be the midpoint of the vertices,
it would be (2,3)
the length of the minor axis is 4, so b = 2
sofar we would would have
(x-2)^2/a^2 + (y-3)^2/4 = 1
but (5,3) lies on it, so
(5-2)^2/a^2 + (3-3)^2/4 = 1
9/a^2 = 1
a^2 = 9
final equation
(x-2)^2/9 + (y-3)^2/4 = 1
thanks!
8(16)
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