原來，2019 M2 卷既 Q9 竟然可以只用 Core 方法做到架，而且考核既知識同 2019 Maths Core Paper I Q19 既差唔多，都係 Quaratics、Transformation of graphs、Coordinate geometry、Mensuration 等，就等我地睇下點做啦！
(a)
Note that $y=\frac{1}{3}\sqrt{12-x^{2}}$ can give $x^{2}+9y^{2}=12$.
On $\Gamma$, when $x=3$, $y=\frac{1}{\sqrt{3}}$.
Therefore, let $y-\frac{1}{\sqrt{3}}=m(x-3)$ be the equation of $L$, where $m$ is the slope of $L$, i.e. $y=m(x-3)+\frac{1}{\sqrt{3}}$.
Then, we have $x^{2}+9\left[m(x-3)+\frac{1}{\sqrt{3}}\right]^2=12$, which gives $\left(9m^{2}+1\right)x^{2}-\left(54m^{2}-6\sqrt{3}m\right)x+\left(81m^{2}-18\sqrt{3}m-9\right)=0$.
Since $L$ is a tangent to $\Gamma$, we know that in the above equation, we know that in the above equation, $\Delta=0$, implying $\left[-\left(54m^{2}-6\sqrt{3}m\right)\right]^{2}-4\left(9m^{2}+1\right) \left(81m^{2}-18\sqrt{3}m-9\right)=0$ so $3m^{2}+2\sqrt{3}+1=0$. Hence, $\left(\sqrt{3}m+1\right)^{2}=0$ and then $m=\frac{1}{\sqrt{3}}$.
Hence, the equation of $L$ is $y-\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}(x-3)$, i.e. $x+\sqrt{3}y-4=0$.
(b)(i)
Note that $y=\sqrt{4-x^{2}}$ gives $x^{2}+y^{2}=4$.
Substitute $y=-\frac{1}{\sqrt{3}}(x-3)+\frac{1}{\sqrt{3}}=-\frac{1}{\sqrt{3}}x+\frac{4}{\sqrt{3}}$ into the above equation, we have $x^{2}+\left(-\frac{1}{\sqrt{3}}x+\frac{4}{\sqrt{3}}\right)^{2}=4$, which implies that $(x-1)^{2}=0$. Hence, $x=1$.
Also, since $1$ is a double root of the equation, we know that $L$ really touches $C$ at $x=1$.
On $C$, when $x=1$, $y=\sqrt{3}$ so the required coordinates are $\left(1,\sqrt{3}\right)$.
(b)(ii)
By solving $x^{2}+9y^{2}=12$ and $x^{2}+y^{2}=4$, we know that $x^{2}=3$ and since $x>0$, we have $x=\sqrt{3}$. Moreover, since $y=\sqrt{4-x^{2}}\geq 0$, we know that $y=1$.
Hence, the required coordinates are $\left(\sqrt{3},1\right)$.
(b)(iii)
Note that the $C$ is in fact a quarter of a circle with centre $(0,0)$ and radius $2$.
Moreover, note that the graph of $y=\sqrt{12-x^{2}}$, where $x\in\left(0,2\sqrt{3}\right)$, denoted by $C'$, is a quarter of a circle with centre $(0,0)$ and radius $2\sqrt{3}$ and this graph can be reduced along the $y$-axis to $\frac{1}{3}$ of the original one to the curve $\Gamma$.
Then, we can use the filling method to find the required area.
Let $O$ be the origin, $P$ be the point $\left(3,\frac{1}{\sqrt{3}}\right)$, $Q$ be the point $\left(1,\sqrt{3}\right)$ and $R$ be the point $\left(\sqrt{3},1\right)$.
First, the area of $\triangle OPQ$ is $(3-0)\left(\sqrt{3}-0\right)-\frac{1}{2}(1-0)\left(\sqrt{3}-0\right)-\frac{1}{2}(3-0)\left(\frac{1}{\sqrt{3}}-0\right)-\frac{1}{2}(3-1)\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) = \frac{4}{3}\sqrt{3}$.
Second, note that if $\alpha$ is the angle between $OQ$ and the $x$-axis, then $\tan\alpha=\frac{\sqrt{3}}{1}$ so $\alpha = 60^{\circ}$ and if $\beta$ is the angle between $OR$ and the $x$-axis, then $\tan\beta=\frac{1}{\sqrt{3}}$ so $\beta=30^{\circ}$ so $\angle QOR=60^{\circ}-30^{\circ}=30^{\circ}$. As a result, the area of the circular sector $OQR$ of $C$ can be given by $\pi (2)^{2}\times\frac{30^{\circ}}{360^{\circ}}=\frac{\pi}{3}$.
Third, note that the region bounded by $\Gamma$, $OP$ and $OR$ can be obtained by reducing a circular sector of the $C'$ along the $y$-axis to $\frac{1}{3}$ of the original one so the area of the former is $\frac{1}{3}$ of the latter. Before the reduction, the coordinates of $P$ and $R$ are $\left(3,\sqrt{3}\right)$ and $\left(\sqrt{3},3\right)$ respectively, denoted by $P'$ and $R'$ respectively. Hence, if $\gamma$ is the angle between $OP'$ and the $x$-axis, then $\tan\gamma=\frac{\sqrt{3}}{3}$ so $\gamma=30^{\circ}$ whereas if $\delta$ is the angle between $OR'$ and the $x$-axis, then $\tan\delta=\frac{3}{\sqrt{3}}$ so $\delta=60^{\circ}$. Hence, $\angle P'OR'=60^{\circ}-30^{\circ}=30^{\circ}$. As a result, the area of this region can be found by $\pi \left(2\sqrt{3}\right)^{2} \times \frac{30^{\circ}}{360^{\circ}} \times \frac{1}{3} = \frac{\pi}{3}$.
Consequently, the required area is $\frac{4}{3}\sqrt{3}-\frac{\pi}{3}-\frac{\pi}{3}=\frac{4}{3}\sqrt{3}-\frac{2\pi}{3}$.
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