The curve is described by the parametric equations $x = \frac{2t + 3}{t + 2}$ and $y = t^2 + at + 1$, where $a$ is a constant. At the point $P$ on the curve, the gradient is given as $1$.
(a)[4]
Show that the value of $t$ at $P$ satisfies the equation $2t^3 + (a + 8)t^2 + (4a + 8)t + 4a - 1 = 0$.
(b)[2]
It is given that $(t + 1)$ is a factor of $2t^3 + (a + 8)t^2 + (4a + 8)t + 4a - 1$. Find the value of $a$.
(c)[3]
Hence show that $P$ is the only point on the curve where the gradient is $1$.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use the quotient rule (or an equivalent approach) in an attempt to find $\frac{dx}{dt}$” …