Why is the equation of normal to the curve is x=10/9?

Question

    I have x = t^2 +1 and y = 3t^2-2t, These are the Parametric Equations. 
    dy/dx = 3-1/t.
    I already let dy/dx = 0
   3-1/t = 0
   3t-1=0
    t=1/3 

   sub t = 1/3 into x and y respectively
     x = (1/3)^2+1
       = 10/9
     y = 3(1/3)^2 - 2(1/3)
       = -1/3

     Coordinates of turning point(10/9,-1/3)

   I don't know how to solve the equation of the curve? According to the solution the equation of 
   normal to the curve is x=10/9. So how can I find the equation of normal to the curve?

Answer 1

By your condition of dy/dx = 0, you have determined that the tangent to the curve is a horizontal line passing through (10/9, -1/3) i.e. y = -1/3.
Thus, the normal (line perpendicular to the tangent) at that point is x = 10/9.
Really, you have completed all the steps necessary beside the final step of interpreting the question.