The slope of the tangent to the curve x=3 t^2+1 , y=t^3-1 at x=1 is:

Step-by-step solution

Given curve is x=3 t^2+1 Second curve is y=t^3-1 d y d x = d y d t d t d x =3 t^2 1 6 t = t 2 But from (i) when x=1 we have 1=3 t^2+1 3 t^2=0 t=0 When x =1 then t =0 dy dx = 0 Hence, slope of the tangent to the curve =0