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.

You can let $C$ have coordinates $(x_1,y_1)$ and $D$ have coordinates $(x_2,y_2)$. 
Hence P has coordinates $(\frac{\lambda x_1+x_2}{\lambda +1},\frac{\lambda y_1+y_2}{\lambda +1})$
Asymptotes have equations $y=3x$ and $y=-3x$, hence $y_1=3x_1$ and $y_2=-3x_2$
Therefore P has coordinates $(\frac{\lambda x_1+x_2}{\lambda +1},\frac{3\lambda x_1-3x_2}{\lambda +1})$. 
P also lies on the hyperbola hence substituting these coordinates into the equation of the hyperbola will get you a relationship between $x_1x_2$ and $\lambda$.
This allows you to find an expression for the area in terms of $\lambda$  which can then be minimised with derivatives.
