find the equation of the normal to the curve x= 2cosALPHA, y = 3sinALPHA at the point where ALPHA= pi/4

find the coordinates of the point where this normal cuts the curve again.

i didnt have any problems with the first question; i found the equation of the normal is y = 2x/3 - 2sqrt2/3 + 3sqrt2/2

but from here its hard to find the point where the curve and the normal intersect again. i tried using 3sinALPHA=2x/3 - 2sqrt2/3 + 3sqrt2/2 and replacing ALPHA with arccos(x/2) but that doesnt make sense does it?