Remember that from your parametric eqn you can get (dy/dx) which is =(dy/dθ) ÷ (dx/dθ). So you know confirm somewhere you need to force out a gradient (also known as tangent) which can be found since the question gives you the gradient of the normal (normal // to y=x). So you derive the gradient of tangent and equate it to dy/dx and solve. Thereafter remember that question asking you for coordinates when the above happens which means need to sub in the θ into the y and x.