$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$

Question

Solve the system: $$\begin{array}{|l} \dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$

First, we have $x,y \ne 0$. Let's write the first equation as: $$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$ We have $x^2-y^2=5$, therefore $xy=6$. What to do next?

Answer 1

Rewrite $\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6}$ as $6x^2-5xy-6y^2=0$ and then factorize,

$$(2x-3y)(3x+2y)=0$$

to have $x=\frac32y$ and $x=-\frac23 y$. Plug them into $x^2-y^2=5$ to obtain the real solutions $(3,2)$ and $(-3,-2)$.

Answer 2

$xy = 6 \implies y= \dfrac6x$ . So ,

$$x^2-y^2 = 5 \implies x^2 - \dfrac{36}{x^2} =5$$

Taking $x^2 = a$ , we get :

$$a^2-36=5a\implies (a-9)(a+4) = 0$$ We have $a=9\implies x=\pm3$ and $y = \pm2$

And we have $a=-4 \implies x = \pm 2i$ and $y=\pm 3i$

So the solutions are $(x,y) = (3,2)$ , $(x,y) = (-3,-2)$ ,

$(2i,-3i)$ and $(x,y) = (-2i,3i)$

Answer 3

The given problem is equivalent to finding the intersections between a rectangular hyperbola and two lines through the origin, since the first equation gives $\frac{y}{x}\in\left\{\frac{2}{3},-\frac{3}{2}\right\}$. These lines are orthogonal, so it is pretty simple to locate the solutions $(3,2)$ and $(-3,-2)$. The line with slope $-\frac{3}{2}$ does not intersect the hyperbola, whose asymptotes are $y=\pm x$.

Answer 4

HINT

From the trigonometric point of view, one can substitute $x = r\cos(\theta)$ and $y = r\sin(\theta)$, from whence we get \begin{align*} \begin{cases} r^{2}\cos(2\theta) = 5\\\\ r^{2}\sin(2\theta) = 12 \end{cases} \Longrightarrow \frac{25}{r^{4}} + \frac{144}{r^{4}} = 1 \Longrightarrow r^{4} = 169 \Longrightarrow r = \sqrt{13} \end{align*}

since $r \geq 0$. Can you take it from here?