A curve is given in parametric form by $x = \frac{2t + 3}{t + 2}$ and $y = t^2 + at + 1$, where $a$ is constant. At the point $P$ on the curve, the gradient is 1.
(a)[4]
Demonstrate that the value of $t$ at $P$ satisfies $2t^3 + (a + 8)t^2 + (4a + 8)t + 4a - 1 = 0$.
(b)[2]
It is stated that $(t + 1)$ is a factor of $2t^3 + (a + 8)t^2 + (4a + 8)t + 4a - 1$. Determine $a$.
(c)[3]
Hence prove 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, to try to find $\dfrac{dx}{dt}$” …