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the centre of this one is (0,0) and the major axis lies horizontally, so
x^2/a^2 + y^2/b^2 = 1 **
for such an ellipse: a^2 = b^2 + c^2, but c = 4
a^2 = b^2 + 16
sub back into **
x^2/(b^2+16) + y^2/b^2 = 1
but (4,1) lies on it, so
16/(b^2+16) + 1/b^2 = 1
16b^2 + b^2+16 = b^2(b^2 + 16)
17b^2 + 16 = b^4 + 16b^2
b^2 = (1 ± √65)/2 = (1 + √65)/2 since b^2 > 0
then a^2 = (33 + √65)/2
x^2 / ( (33 + √65)/2) + y^2/( (1 + √65)/2) = 1