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I'm currently investigating the curve implicitly defined by $x^2-y^2= \ln (xy)$. Now I can see that by considering the region $xy>1$ and $xy<1$, we can determine where the curve is above or below the line $y=x$.

However, I do not feel that this is not enough to prove that the line $y=x$ is an asymptote in itself, just whether it approaches it from above or below.

For an implicitly defined curve like this, what kind of approach would I take to show that $y=x$ is an asymptote?

Trogdor
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  • Have you seen this wiki page? https://en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes – Berci Sep 20 '15 at 10:03
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    Yes I have. My issue is with the fact that there are two variables here, so I cannot simply take the limit as $x\rightarrow \infty$ and assume $y$ to be constant. – Trogdor Sep 20 '15 at 10:24
  • In fact $y\to\infty$ as $x\to\infty$. Did you already know that? – David K Sep 20 '15 at 13:32

2 Answers2

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You can establish that $$\frac{dy}{dx}=\frac{2x-\frac 1x}{2y+\frac 1y}\rightarrow 1$$ as $x,y\rightarrow +\infty$

Hence for large $x,y>0$, the graph is approximated by the line $y=x$

David Quinn
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You could separate the variables:

$$x^2 - \ln x = y^2 + \ln y.$$

It's easy then to show that $y\to\infty$ as $x\to\infty$.

Take the ratio of the two sides: $$\frac{y^2 + \ln y}{x^2 - \ln x} = 1.$$

For large positive $x$, you have large positive $y$ and $y^2/x^2 \approx 1$, so $y/x \approx 1$.

You can even write \begin{align} \frac{y^2}{x^2} & = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right) \frac{y^2 + \ln y}{x^2 - \ln x} \\ & = \left( \frac{y^2}{y^2 + \ln y} \right) \left( \frac{x^2 - \ln x}{x^2}\right) \end{align}

Each of the quantities in parentheses goes to $1$ as $x\to\infty$.

David K
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