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?
 A: Part (i)

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.   

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.
A: 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 :
$$7+4\sqrt{3}=4+4\sqrt{3}+3=(2+\sqrt{3})^2$$
