The curve C has parametric equations

x = 4 cos 2t, y = 3 sin t, -pi/2 < t < pi/2

A is the point (2, 3/2), and lies on C.

a) Show that an equation of the normal to C at A is 6y – 16x + 23 = 0.

The normal at A cuts C again at the point B.

b) Find the y-coordinate of the point B.

I’ve done a, but how can I find b. Please help me.