Hi:
I have done part a for this parametric equation question. Before I continue to part b could somebody please check I have succeeded with part a.
thank you
You have done part (a) correctly. For part (b), you are given that the slope of the normal to the curve is -1/2, which means that the slope of the tangent (which is equal to the value of the derivative) at that point is 2.
Therefore, solve for t:
$\displaystyle \frac{2(6e^{4t} - 1)}{18e^{2t} - 1} = 2$