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, then write the line equation. While it involves trigonometric identities and some algebraic manipulation, it follows a well-practiced procedure with no novel insights required, making it slightly easier than average.
4 A curve is defined by the parametric equations
$$x = 4 \cos ^ { 2 } t , \quad y = \sqrt { 3 } \sin 2 t ,$$
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 \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-06_2718_35_141_2012}
Obtain forms \(\frac{dx}{dt} = k_1\cos t\sin t\) or \(\frac{dy}{dt} = k_2\cos 2t\)
*M1
Obtain correct \(-8\cos t\sin t\) or \(-4\sin 2t\) and \(2\sqrt{3}\cos 2t\)
A1
Attempt value of \(\frac{dy}{dx}\) when \(t = \frac{1}{6}\pi\)
*DM1
Need to see attempt at substitution
Obtain \(\frac{dy}{dx} = -\frac{1}{2}\)
A1
State or imply gradient of normal is \(2\)
**M1 FT
Following *their* value of the first derivative
Attempt equation of normal
**DM1
Not tangent and with attempt to find coordinates \(\left(3, \frac{3}{2}\right)\)
Obtain \(4x - 2y - 9 = 0\)
A1
Or equivalent of requested form
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Obtain forms $\frac{dx}{dt} = k_1\cos t\sin t$ or $\frac{dy}{dt} = k_2\cos 2t$ | *M1 | |
| Obtain correct $-8\cos t\sin t$ or $-4\sin 2t$ and $2\sqrt{3}\cos 2t$ | A1 | |
| Attempt value of $\frac{dy}{dx}$ when $t = \frac{1}{6}\pi$ | *DM1 | Need to see attempt at substitution |
| Obtain $\frac{dy}{dx} = -\frac{1}{2}$ | A1 | |
| State or imply gradient of normal is $2$ | **M1 FT | Following *their* value of the first derivative |
| Attempt equation of normal | **DM1 | Not tangent and with attempt to find coordinates $\left(3, \frac{3}{2}\right)$ |
| Obtain $4x - 2y - 9 = 0$ | A1 | Or equivalent of requested form |
4 A curve is defined by the parametric equations
$$x = 4 \cos ^ { 2 } t , \quad y = \sqrt { 3 } \sin 2 t ,$$
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 $a x + b y + c = 0$ where $a , b$ and $c$ are integers.\\
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-06_2718_35_141_2012}\\
\hfill \mbox{\textit{CAIE P2 2024 Q4 [7]}}