The curve $C$ is defined by the parametric equations $x = 6t - 5$ and $y = 2t^2 + 1$, where \(

Question

The curve $C$ is defined by the parametric equations $x = 6t - 5$ and $y = 2t^2 + 1$, where $t \ge \frac{5}{6}$. Sketch the curve $C$, labelling the exact coordinates of any axial intercept(s).[1] Find the equation of the normal to $C$ at the point $P$ where $t = 1$.[2] Find the acute angle between the normal to $C$ at the point $P$ where $t = 1$ and the tangent to $C$ at the point $M$ where $t = 3$.[2] The curve $C$ is translated $5$ units in the positive $x$-direction and translated $3$ units in the negative $y$-direction, to form the curve $D$. Find the equation of $D$ in parametric form.[2] The curve $E$ is defined by the parametric equations $x = 7u$ and $y = \frac{9}{u}$, where $u \neq 0$. Show that at the point of intersection of the curves $C$ and $E$, $6t^3 - 5t^2 + 3t - 34 = 0$. Deduce that there is only one point of intersection and find the coordinates of this point.[4]

Accepted Answer

(a) Closed endpoint $\left(0,\frac{43}{18}\right)$, $x\ge0$; no $x$-intercept. (b) $y=-\frac32x+\frac92$. (c) $60.3^\circ$. (d) $x=6t$, $y=2t^2-2$, $t\ge\frac56$. (e) Cubic shown; unique intersection $(7,9)$.

2025 RVHS P1 Q12

2025 RVHS P1 Q12

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