OK Lor:Hi,Pls help on (ii):A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).(i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.(ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.Ans: -4/3Thanks.Assuming (i) has been proven.tangent at t=3 has equation [(3^2+1)^2]y+3x=3*3^2-3+1simplified, we have 100y+3x=25to find the point of intersection between the curve and the straight line, we need to solve100y+3x=25 and x=2t-1, y=1/(t^2+1)By subsititution,100/(t^2+1)+3(2t-1)=25100+(6t-3)(t^2+1)=25(t^2+1)100+6t^3-3t^2+6t-3=25t^2+256t^3-28t^2+6t+72=03t^3-14t^2+3t+36=0since we know t=3 is a repeated root as a tangent, (t-3)^2 must be a factor, (or t^2-6t+9 must be a factor)after doing a long division, we have(3t+4)(t-3)^2=0the other root is t=-4/3 (when 6t+8=0)