More Golden ratio - Link to $x^2 - y^2 =xy $

Question

So, I like to mess around on Desmos. And there I am, looking at my all old graphs and I came across this one ($x^2-y^2=xy$) :

And now that I know calculus (this was when I was $11$), I got all interested in that gradient (I will focus on the line in the first quadrant). And since it looked linear, I checked if it intersected at any integer values (to make finding the gradient easier)

And guess what. The line almost perfectly intersected at consecutive Fibonacci number values for $x$ and $y$. For example:

When $x=8, y \approx 5$

$x=21, y \approx 13$

$x=55, y \approx 34$

(Try some for yourself)

This leads me to make the assumption $y =\frac{x}{\phi}$, hence the gradient is $\frac{1}{\phi}$.

However, working out the derivative (through implicit differentiation) gives me a different answer:

$x^2-y^2=xy$

$2xdx-2ydy=xdy+ydx$

$2xdx-ydx=xdy-2ydy$

$dx(2x-y)=dy(x-2y)$

$\frac{2x-y}{x-2y}=\frac{dy}{dx}$

This is not the answer of phi that I was expecting - What's going on, can I extract the goden ratio out of this, and what link is there between the golden ratio and the graph?

Answer 1

No need for calculus. One can rewrite the original equation as $$x^2-xy-y^2=(x-\varphi y)\left(x+\frac{y}{\varphi}\right)=0.$$ Therefore, the graph is just the union of the lines with slopes $\frac{1}{\varphi}$ and $-\varphi$ through the origin.

Answer 2

For each fixed $x$, you can solve the equation for $y$ giving $$ y=-\varphi x \vee y=\frac1\varphi x $$

Answer 3

Note that

$$ x^2 - y^2 = xy \Longrightarrow \left(\frac{y}{x}\right)^2 +\frac{y}{x} -1 = 0\Longrightarrow \frac{y}{x}= \frac{-1\pm\sqrt{5}}{2} = \frac{1}{\phi}\lor-\phi. $$ So the slopes of the lines are indeed the golden ratio and its inverse.