Use the formula in Exercise 42 to find the curvature. x=acos(ωt), y=bsin(ωt)

Question

Exercise 42 formula: k = l(x' y'' - y' x'')l / (x'^2 + y'^2)^(3/2)
This is what I did but don't know why my answer not correct.
x'=-aωsin(t), y'=bωcos(ωt),x''=-aω^2cos(ωt), y''=-bω^2sin(ωt)
l[(-aωsin(ωt)*-bω^2sin(ωt)) - (bωcos(ωt)*-aω^2cos(ωt))]l/[(-aωsin(ωt))^2 + (bωcos(ωt))^2]^3/2
l[(abω^3sin^2(ωt)+baω^3cos^2(ωt)]l/[(-a^2ω^2sin^2(ωt) + b^2ω^2cos^2(ωt))]^3/2
labllω^3l/(ω^2)^3/2[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
labllω^3l/ω^3[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
my answer
labl/[-a^2(sin^2(ωt)) + b^2(cos^2(ωt))]^3/2
Why my answer wrong and where did I make mistake at?

Answer 1

you should not have any minus signs in the denominator, since everything is squared.