Find the equations of the tangent and normal to the curve

Find the equations of the tangent and normal to the curve $x(x-y)+1=0$ at an arbitrary point, and determine the limiting position of the tangent as the point of tangency recedes to infinity.

Step 1: An explicit form

Expanding, $x^{2}-xy+1=0 \Rightarrow xy = x^{2}+1$, so

\[ y = x+\frac1x, \qquad x\neq0. \tag{1} \]

Step 2: Derivative

\[ y' = 1-\frac{1}{x^{2}}. \tag{2} \]

Step 3: Tangent at an arbitrary point

At $P=\left(x_{0},\,x_{0}+\dfrac1{x_{0}}\right)$ the slope is $k = 1-\dfrac1{x_{0}^{2}}$, and

\[ y - \left(x_{0}+\frac1{x_{0}}\right) = \left(1-\frac1{x_{0}^{2}}\right)(x-x_{0}). \tag{3} \]

Expanding the right side (the terms $-x_{0}+\tfrac1{x_0}+x_0+\tfrac1{x_0}$ collapse to $\tfrac2{x_0}$): \[\text{Tangent: } \quad y = \left(1-\frac{1}{x_{0}^{2}}\right)x + \frac{2}{x_{0}}. \tag{4}\]

Step 4: Normal at the same point

The slope of the normal is $m = -\dfrac1k = -\dfrac{x_{0}^{2}}{x_{0}^{2}-1}$ (for $x_{0}\neq\pm1$). Writing the normal and combining the constant terms over the denominator $x_{0}(x_{0}^{2}-1)$,

\[ \frac{x_{0}^{3}}{x_{0}^{2}-1}+x_{0}+\frac1{x_{0}} = \frac{x_{0}^{4}+x_{0}^{2}(x_{0}^{2}-1)+(x_{0}^{2}-1)}{x_{0}(x_{0}^{2}-1)} = \frac{2x_{0}^{4}-1}{x_{0}(x_{0}^{2}-1)}, \]

\[\text{Normal: } \quad y = -\frac{x_{0}^{2}}{x_{0}^{2}-1}\,x + \frac{2x_{0}^{4}-1}{x_{0}(x_{0}^{2}-1)}. \tag{5}\]

Step 5: Limiting position as $x_{0}\to\infty$

In (4), $\displaystyle\lim_{x_0\to\infty}\left(1-\frac{1}{x_0^2}\right)=1$ and $\displaystyle\lim_{x_0\to\infty}\frac{2}{x_0}=0$, so the tangent approaches

\[ y = x. \tag{6} \]

The same limit is obtained as $x_{0}\to-\infty$.

Step 6: Geometric interpretation

The line $y=x$ is in fact an asymptote of the curve, since $y-x = 1/x \to 0$ as $x\to\pm\infty$. The limiting position of the tangent, as the point of tangency goes to infinity, coincides with the asymptote — a general phenomenon for curves with an oblique asymptote.

Step 7: Verification

\[\text{Tangent (4) at } x=x_0\text{:}\quad \left(1-\tfrac1{x_0^2}\right)x_0+\tfrac2{x_0}=x_0-\tfrac1{x_0}+\tfrac2{x_0}=x_0+\tfrac1{x_0}=y_0\ \checkmark\]

\[\text{Asymptote check: } y-x=1/x\to0 \text{ as } x\to\infty\ \checkmark\]