The curve is given in parametric form by $x = t^2 + 1$, $y = 4t + \ln(2t - 1)$.
(i)[3]
Express $\frac{dy}{dx}$ as a function of $t$.
(ii)[3]
Find the equation of the normal to the curve at the point where $t = 1$. Present your answer in the form $ax + by + c = 0$.
(c(ii))[3]
Find the equation of the normal to the curve at the point where $t = 1$. Present your answer in the form $ax + by + c = 0$.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State that $\dfrac{dy}{dt}=4+\dfrac{2}{2t-1}$” …
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