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

Question

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.

Answer 1

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.

Answer 2

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. $\color{red}{\lambda=1}$.