Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: find dx/dt and dy/dt, compute dy/dx = (dy/dt)/(dx/dt), evaluate at the given parameter value, find the normal gradient, and write the line equation. While it involves multiple steps and trigonometric manipulation, it follows a completely standard algorithm with no novel insight required, making it slightly easier than average.
A curve is defined by the parametric equations
$$x = 4\cos^2 t, \quad y = \sqrt{3}\sin 2t,$$
for values of \(t\) such that \(0 < t < \frac{1}{2}\pi\).
Find the equation of the normal to the curve at the point for which \(t = \frac{1}{6}\pi\). Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers. [7]
Not tangent and with attempt to find coordinates 3, .
2
Answer
Marks
Guidance
Obtain 4x2y90
A1
Or equivalent of requested form.
7
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | dx dy
Obtain forms k costsint or k cos2t
dt 1 dt 2 | *M1
Obtain correct 8costsint or 4sin2t and 2 3cos2t | A1
dy
Attempt value of when t 1π
dx 6 | *DM1 | Need to see attempt at substitution.
dy 1
Obtain
dx 2 | A1
State or imply gradient of normal is 2 | **M1FT | Following their value of the first derivative.
Attempt equation of normal | **DM1 | 3
Not tangent and with attempt to find coordinates 3, .
2
Obtain 4x2y90 | A1 | Or equivalent of requested form.
7
Question | Answer | Marks | Guidance
A curve is defined by the parametric equations
$$x = 4\cos^2 t, \quad y = \sqrt{3}\sin 2t,$$
for values of $t$ such that $0 < t < \frac{1}{2}\pi$.
Find the equation of the normal to the curve at the point for which $t = \frac{1}{6}\pi$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$ and $c$ are integers. [7]
\hfill \mbox{\textit{CAIE P2 2024 Q4 [7]}}