Find the value of $\lambda$ that maximises the area of a triangle on a hyperbola. 

Consider the hyperbola: $$ \frac{x^2}{4} - \frac{y^2}{36}=1$$
$l_1$ and $l_2$ are the asymptotes

Consider a case where $P$ is located in the first quadrant. Through $P$, draw another line $CD$, where $C$  is on line $l_1$, D is on line $l_2$, and $P$ is between $C$ and $D$, such that $CP:PD=1: \lambda $
Find the value of $\lambda$, which will minimise the area of the triangle $COD$, and find this area.


What I have done: 
The asymptotes of the graph are $y=3x$ and $y=-3x$
The point $P$ lies on the hyperbola so let the coordinates of this point be $(x_0 , y_0)$ with the properties that $x_0 >0$ and hence $ \frac{x_0^2}{4}-\frac{y_0^2}{36}=1$ which implies that $x_0 = \frac{\sqrt{36+y_0^2}}{3}$
The point $D$ is determined by $C$ and $P$ , so let's find C, a point $(c_1,c_2)$ with the properties that $c_1>0$ and $c_2=3c_1$
This is where I am stuck , I don't know if my approach is correct and even if it is I feel like the algebra is too complex , there should be a more simpler way of doing this.

Rishi has already provided a good answer.
This answer shows the detail on how to find the coordinates of $P$ and a way to find the minimum area of $\triangle{COD}$ without using derivatives.
Let $C(x_1,3x_1),D(x_2,-3x_2)$. 
Since $CP:PD=1:\lambda$ leads that $\vec{CP}=\frac{1}{\lambda+1}\vec{CD}$, we get $$\vec p=\vec c+\frac{1}{\lambda+1}(\vec d-\vec c)=\frac{\lambda \vec c+\vec d}{\lambda+1}$$to have
$P(\frac{\lambda x_1+x_2}{\lambda+1},\frac{3x_1\lambda-3x_2}{\lambda+1})$.
Since $P$ is on the hyperbola, we get
$$\frac 14\left(\frac{\lambda x_1+x_2}{\lambda+1}\right)^2-\frac{1}{36}\left(\frac{3x_1\lambda-3x_2}{\lambda+1}\right)^2=1,$$
i.e.
$$x_1x_2=\frac{(\lambda+1)^2}{\lambda}\tag1$$
Also, letting $\alpha=\angle{COD}$, we get that 
$$[\triangle{COD}]=\frac 12\times\overline{OC}\times\overline{OD}\times\sin{\alpha}=\frac 12\sqrt{x_1^2+9x_1^2}\sqrt{x_2^2+9x_2^2}\times\frac 35=3x_1x_2\tag2$$
where $\sin{\alpha}=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}=2\times\frac{3}{\sqrt{10}}\times\frac{1}{\sqrt{10}}=\frac 35$ since $\tan\frac{\alpha}{2}=3$.
From $(1)(2)$, we get, by the AM-GM inequality,
$$[\triangle{COD}]=3\times\frac{(\lambda+1)^2}{\lambda}=3\left(\lambda+\frac{1}{\lambda}+2\right)\ge 3\left(2+2\right)=12$$
Therefore, the minimum area of $\triangle{COD}$ is $\color{red}{12}$ when $\lambda=\frac{1}{\lambda}$, i.e. $\color{red}{\lambda=1}$.
