1

Consider the hyperbola: $$ \frac{x^2}{4} - \frac{y^2}{36}=1$$

<p>$l_1$ and $l_2$ are the asymptotes</p>

<p><a href="https://i.stack.imgur.com/5fqUV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5fqUV.jpg" alt="enter image description here"></a></p>

<p>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 $</p>

<p>Find the value of $\lambda$, which will minimise the area of the triangle $COD$, and find this area.</p>

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.

bigfocalchord
  • 5,597

2 Answers2

4

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

mathlove
  • 139,939
  • Thank you for the answer but I was wondering if there was anyway I could continue with my method? – bigfocalchord Dec 15 '16 at 07:04
  • I also don't understand: Since $CP:PD=1:\lambda$ leads that $\vec{CP}=\frac{1}{\lambda+1}\vec{CD}$ – bigfocalchord Dec 15 '16 at 07:05
  • @dydxx: To the first comment : Yes, you can, but anyway you will need to use the formula for internal division of a line segment. If you don't want to use the formula, then I think you have to have very tedious calculations. To the second : $P$ is on the line segment $CD$ with $CP:PD=1:\lambda$. So, there is a positive real number $k$ such that $|CP|=k,|PD|=k\lambda,|CD|=|CP|+|PD|=k(\lambda+1)$. So, $|CP|=\frac{1}{\lambda+1}|CD|$ holds and so $\vec{CP}=\frac{1}{\lambda+1}\vec{CD}$ holds. Drawing the line segment $CD$ should help. – mathlove Dec 15 '16 at 07:37
  • Thank you for the comment, to the first comment of what you said , I do not want to use the internal division formula and I want to continue on with my method. If you could show me how in another comment I will gladly give you the +50. :) – bigfocalchord Dec 15 '16 at 07:49
  • @dydxx: Well, you already have $C(c_1,3c_1),P(\sqrt{36+y_0^2}/3,y_0)$. You can determine $D$ by $C,P$ since $D$ is the intersection point of $y=-3x$ with the line $CP$. Now, representing $CP^2$ and $PD^2$ by $y_0,c_1$ and using $CP:PD=1:\lambda$, i.e. $\lambda^2\times CP^2=PD^2$ will give you a relation between $y_0,c_1,\lambda$. Then, you will be able to represent the area of $\triangle{COD}$ by $\lambda$, and done. – mathlove Dec 15 '16 at 08:05
  • @dydxx: My suggestion might not work. In other words, your method without using internal division formula might not work. Let $D(d,-3d)$. Then, we can determine $d$ by $y_0,c_1$ and get $[\triangle{COD}]=3cd$. However, I found it very difficult (maybe impossible) to represent $3cd$ by $\lambda$ using $\lambda^2\times CP^2=PD^2$ instead of $CP:PD=1:\lambda$. So, if we don't use $CP:PD=1:\lambda$ itself (i.e. the formula itself), we might not be able to find the answer. This is all I can say now. I hope this helps. – mathlove Dec 16 '16 at 05:39
  • Hi I will accept your answer , however do you know of any other method of solving this question? – bigfocalchord Dec 19 '16 at 06:11
  • 1
    @dydxx: Thanks. Well, sorry but I have no idea. From the condition $CP:PD=1:\lambda$, the internal division formula gives us two pieces of information, i.e. the information for $x$ and information for $y$. However, if we don't use the formula, I think that we have to use the condition in the form $\lambda^2\times CP^2=PD^2$, i.e. the relation of the lengths of some line segments. This form of the condition gives us only one piece of information, i.e. we can't separate it into $x$ and $y$. So, I think that the key is to use the internal division formula. – mathlove Dec 19 '16 at 06:21
  • It's just that this question was given and I had no clue what the internal division formula was so I was extremely confused how to start solving this question! So I posted it here that maybe there would be an answer without using it. Anyways thank you for your comprehensive answer as always. Take care. – bigfocalchord Dec 19 '16 at 06:23
  • @dydxx: I see, and you are welcome :) – mathlove Dec 19 '16 at 06:25
2

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.

Rishi
  • 833