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Question: Figure $3$ shows six lines passing through the origin. The lines are separated by equal angles. Some exact values of $\tan(t)$ are given in Table $1$.

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

<p>$(i)$ Show that the lines  can be represented by the following equation:</p>

<p>$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$</p>

<p>$(ii)$ Find an equation for a hyperbola that does not cross any of the six lines in Figure $3$, giving
reasons for your answer.</p>

I'm just really stuck , how do I even start this question! My approach has been this:

Let $y=mx+c$ for an equation of any line since all lines pass through the origin $(0,0)$ then $y=mx$ and because $m=\tan(t)$ we have the equation of any of these lines is $y=\tan(t)x$

And since there are $6$ lines passing through the origin there are $12$ sub divisions which means the graph is separated into $12$ parts and the angle between each of the parts will be $\frac{2\pi}{12}=\frac{\pi}{6}$

But I am confused , how should I continue? Am I even on the right track?

bigfocalchord
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  • There is no clue that the plane is equally divided by those lines – H.C. Lefevre Oct 26 '16 at 08:24
  • @H.C.Lefevre That was my thinking atleast , I'm trying to figure out what is correct :// – bigfocalchord Oct 26 '16 at 08:24
  • See my edit of the post – H.C. Lefevre Oct 26 '16 at 08:28
  • @H.C.Lefevre Okay sorry but I think you might be misunderstanding me a little bit , my question is that if we weren't given that equations to prove and the question was find equations of 6 lines shown then how would I do that just from the picture and the angles given.. – bigfocalchord Oct 26 '16 at 08:30
  • It appears that the plane is equally divided by those lines, The square root expression is confusing as you can see in the edit of my post – H.C. Lefevre Oct 26 '16 at 08:32
  • No I understand the square root expression and how I can get the angles by working backwards , I just want to know how do I get those angles without working backwards as in by interpreting the graph and finding the angles manually then finding the equations of the lines by subbing into $y=\tan(t) \cdot x$ – bigfocalchord Oct 26 '16 at 08:34
  • The angles in the table don’t all correspond to the slopes of the lines. None of the lines has a slope of $\tan\frac\pi6$ or $\tan\frac\pi3$. – amd Oct 26 '16 at 08:44
  • @amd Yes only those two don't have relevance the other 3 do however – bigfocalchord Oct 26 '16 at 08:45
  • So, what is it that you’re trying to figure out? With only the graph and description of it, you have to guess the slope of one of those lines. Once you’ve settled on a value for that angle, the rest come from “equally spaced.” – amd Oct 26 '16 at 08:49
  • @amd Why do we have to guess surely there is a logical way of coming up with the right angle? – bigfocalchord Oct 26 '16 at 09:00
  • I do not understand your point : It is written that you have six lines separated by equal angles as you shown the angles is $\pi/6$. You have that the lines are crossing the origin, then their equation has the form : $$y=\tan(\theta_n)x$$. You can see that you have $y=x$ (even if there is no metric on the axes, just looking at the angles and supposing that you have an orthonormal frame). So for sure you have $\pi/4$ among those angles. Then you can build up the $\theta_n$ set like this : $$\theta_n=\pi/4+n\times\pi/6$$. – H.C. Lefevre Oct 26 '16 at 09:44
  • You know that the angle between adjacent pairs of lines must be $\pi/6$, but you have no other information that will let you determine any of the angles for sure. The picture certainly suggests that one of the lines has a slope of 1 and that the arrangement is symmetric w/r to the coordinate axes, but there’s no way to know this for sure. For the same reason, the original problem is also not stated well. The given equation does indeed describe a set of six equally-spaced lines, but you’re left to infer that their slopes do in fact match those of the pictured lines. – amd Oct 26 '16 at 10:00
  • just do $y = x \tan \theta $ – Anonymous Oct 30 '16 at 17:12

2 Answers2

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Part (i)

enter image description here

See diagram above. Equations are:

$$\begin{align}\color{red}{x=\pm y \qquad\Rightarrow x^2-y^2=0}\\ \color{blue}{x=\pm y\tan\frac{\pi}{12} \qquad \Rightarrow x^2-y^2\tan^2\frac{\pi}{12}=0}\\ \color{green}{y=\pm x\tan\frac{\pi}{12} \qquad\Rightarrow x^2-y^2\big / \tan^2\frac{\pi}{12}=0} \end{align}$$ Multiplying the three equations and taking note that $$\tan\frac{\pi}{12}=2-\sqrt{3}\\ \tan^2\frac{\pi}{12}=7-4\sqrt{3}\\ \frac 1{\tan^2\frac{\pi}{12}}=7+4\sqrt{3}$$ we have: $$\begin{align} \color{red}{\bigg(x^2-y^2\bigg)} \color{blue}{\bigg(x^2-y^2\tan^2\frac{\pi}{12}\bigg)} \color{green}{\bigg(x^2-y^2\big / \tan^2\frac{\pi}{12}\bigg)}&=0\\ \color{red}{\bigg(x^2-y^2\bigg)} \color{blue}{\bigg(x^2-(7-4\sqrt{3})y^2\bigg)} \color{green}{\bigg(x^2-(7+4\sqrt{3})y^2\bigg)}&=0\quad\blacksquare\\ \end{align}$$

Part (ii)

A family of three hyperbola pairs (in alternating spaces between asymptotes) which do not cross any of the six lines is given by setting LHS of the above equation to a constant, i.e. $$ \bigg(x^2-y^2\bigg) \bigg(x^2-(7-4\sqrt{3})y^2\bigg) \bigg(x^2-(7+4\sqrt{3})y^2\bigg) =c $$ where $c$ is a constant.

enter image description here

Changing the sign of $c$ places the family of hyperbolas in the alternate set of inter-asymptote spaces.

For only one pair of hyperbola, try, e.g.

$$\bigg(x-y\bigg)\left(x-\frac y{\tan(\pi/12)}\right)=-d$$

where $d$ is a constant.

Hypergeometricx
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First you can see that there are six lines on your graph and the given equation is the product of three terms in $x^2$ and $y^2$. So each of those three terms gives two of those lines : $$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$ Is equivalent to : $$x^2-y^2=0$$ $$x^2-(7-4\sqrt{3})y^2=0$$ $$x^2-(7+4\sqrt{3})y^2=0$$ Then separating $x$ and $y$ : $$x^2=y^2\rightarrow y=\pm x$$ $$x^2=(7-4\sqrt{3})y^2\rightarrow y=\pm x\frac{1}{\sqrt{7-4\sqrt{3}}}$$ $$x^2=(7+4\sqrt{3})y^2\rightarrow y=\pm x\frac{1}{\sqrt{7+4\sqrt{3}}}$$

It appears that what confuses you is this, the plane is actually equally divided by those lines: $$\frac{1}{\sqrt{7+4\sqrt{3}}} = 2-\sqrt{3}$$ It comes from : $$7+4\sqrt{3}=4+4\sqrt{3}+3=(2+\sqrt{3})^2$$

H.C. Lefevre
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  • Yes I do understand that part but I want to know how they got those lines , I can work backwards and find the angle that way but I want to know how they got those angles in the first place! – bigfocalchord Oct 26 '16 at 08:18