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

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?

Answer 1

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

Answer 2

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: $$\begin{align} \color{red}{\bigg(x^2-y^2\bigg)} \color{blue}{\bigg(x^2-y^2\tan^2\frac{\pi}{12}\bigg)} \color{green}{\bigg(x^2-y^2\big / \tan^2\frac{\pi}{12}\bigg)}&=0\\ \color{red}{\bigg(x^2-y^2\bigg)} \color{blue}{\bigg(x^2-(7-4\sqrt{3})y^2\bigg)} \color{green}{\bigg(x^2-(7+4\sqrt{3})y^2\bigg)}&=0\quad\blacksquare\\ \end{align}$$

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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.