<s>[INST] Show that the 6 lines can be represented by the following equation: $(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$ 

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 $1$.

$(i)$ Show that the lines  can be represented by the following equation:
$$(x^2-y^2)(x^2-(7-4\sqrt{3})y^2)(x^2-(7+4\sqrt{3})y^2)=0$$
$(ii)$ Find an equation for a hyperbola that does not cross any of the six lines in Figure $3$, giving
    reasons for your answer.


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?
 [/INST] 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 :
$$7+4\sqrt{3}=4+4\sqrt{3}+3=(2+\sqrt{3})^2$$
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