x^2 + 2y^2 =3
find the tangent and normal lines at (1,1)
so I think that I find y' first
2x + 4yy' = 0
y' = (3-2x)/4y
setting this to 0 to get the tangent line slope gives
3-2x=0 and x = 3/2
I think this is so far correct, but I am not sure what isnext.