A curve $C$ is defined by the parametric equations \[x = 4\cos t + 3 \quad \text{and} \quad y = 2\

Question

A curve $C$ is defined by the parametric equations $$x = 4\cos t + 3 \quad \text{and} \quad y = 2\sin t + 1, \quad \text{where}\hspace{0.5em} 0 \le t \le \pi.$$ Sketch the curve $C$, labelling the coordinates of the end points.[2] The point $P$ on $C$ has parameter $p$. Show that the equation of the normal at point $P$ can be expressed in the form $$y = a(\tan p)x + b(\sin p) + c(\tan p) + 1,$$ where $a$, $b$ and $c$ are constants to be determined.[5] Show that the cartesian equation of curve $C$ is $y = 1 + \sqrt{4 - \frac{(x - 3)^2}{4}}$.[4] Using the results from parts (ii) and (iii), find the area of the region bounded by the curve $C$, the line $x = 0$ and the normal to the curve $C$ at the point where $t = 2$, giving your answer correct to 3 decimal places.[3]

Accepted Answer

(ii) ${a = 2,\ b = -6,\ c = -6}$ (iii) Since ${0 \leq t \leq \pi}$ and ${y = 2\sin t + 1}$, ${y > 1}$. ${y = 1 + \sqrt{\left(4 - \frac{(x-3)^2}{4}ight)}}$ $\left( 	ext{Reject } y = 1 - \sqrt{\left(4 - \frac{(x-3)^2}{4}ight)} ight)$

2024 MI PRU2 P2 Q6

2024 MI PRU2 P2 Q6