0

This seems like a very simple computation, but I don't fully understand it.

Consider the line $y = 1/x$ in the first quadrant (i.e., when $x \geq 0$). We fix a point $x_0$ and consider the line tangent to $y = 1/x$ at the point $x_0$. Computationally, it's easier to see that the $y$-intercept is $2x_0$. The claim that I do not understand is that it should be true "by mirror symmetry" that the $x$-intercept immediately is $2y_0$. The argument the professor made is that $$y = 1/x \iff xy = 1 \iff x = \frac{1}{y}.$$ I understand this, but we're looking at the tangent line, not at the original function $y$. I can observe the function and "see" that there may be some mirror symmetry, but I don't understand fully how the geometry works.

user861776
  • 1,062

2 Answers2

1

The original equation $y = (1/x)$ is a hyperbola.

This implies that $x = (1/y).$

This means that instead of regarding $y = f(x) = (1/x)$, you can regard $x = g(y) = (1/y).$

Then all of your analysis, namely that the slope is $\frac{-1}{x^2}$ and that $\frac{1}{x_0} + \frac{x_0}{(x_0)^2} = \frac{2}{x_0}$ will be perfectly paralleled when you consider $x = g(y) = (1/y)$.

user2661923
  • 35,619
  • 3
  • 17
  • 39
1

The tangent at the point $(x_0,{1 \over x_0}) = ({1 \over y_0}, y_0)$ has the equation $y={1 \over x_0} -{1 \over x_0^2}(x-x_0)$, or, viewing it from the $y$ axis perspective $x={1 \over y_0} -{1 \over y_0^2}(y-y_0)$.

It is clear that the equations are the same if we swap the symbol $x$ with $y$ (including the subscripted symbols), however I would not call it mirror symmetry, more

Since the $y$ intercept is found by setting $x=0$ in the first equation (which gives ${2 \over x_0})$, using the $x-y$ swap we see that the $x$ intercept is ${2 \over y_0}$.

copper.hat
  • 172,524