Last night, a netizen discussed the following problem with me:
The graph of the function y=\sqrt{3} x-\frac{1}{x} is a hyperbola. Take points P and Q on the two branches of this hyperbola respectively. Find the shortest distance PQ.
Obviously, if the hyperbola were in the standard form \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, this problem would be quite simple. It would be the distance between the two points when y=0, which is 2a. However, it is clear that this hyperbola has been rotated. Therefore, we need to know exactly how many degrees \theta it has been rotated. Then, by setting y=(\tan\theta) x and solving it simultaneously with the hyperbola equation, we can find the two points.
To find the rotation angle, we need to leverage the two asymptotes of the hyperbola. Problems involving straight lines are always easier to handle than those involving hyperbolas. “Asymptotic” means the approximate shape of the function as it approaches a limit. As x\to \infty, \frac{1}{x} \to 0, meaning one of the asymptotes is y=\sqrt{3} x. Additionally, as x\to 0, y\to \infty, which indicates that the other asymptote is x=0. The situation for the latter asymptote can be understood by analogy with the asymptotes of y=\frac{1}{x}.
We plot the hyperbola and its asymptotes as follows:
Next, we only need to find the equation of the angle bisector of the two asymptotes, which is the green dashed line in the figure. Through simple reasoning, it can be concluded that the angle between the green dashed line and the negative x-axis is 15^\circ, i.e., \frac{\pi}{12}. Using the identity: \frac{2 \tan\frac{\pi}{12}}{1-\tan^2 \frac{\pi}{12}}=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}
We can find that \tan\frac{\pi}{12}=2-\sqrt{3}. Therefore, the equation of the green dashed line is: y=(\sqrt{3}-2)x
Solving this simultaneously with y=\sqrt{3} x-\frac{1}{x}, we get: x=\pm \frac{\sqrt{2}}{2} Correspondingly: y=\pm \left(\frac{\sqrt{6}}{2}-\sqrt{2}\right)
So the distance is: \begin{aligned} 2\sqrt{\left(\frac{\sqrt{6}}{2}-\sqrt{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)} \\ =2\sqrt{4-2\sqrt{3}} \\ =2(\sqrt{3}-1) \end{aligned}
Reprinting please include the original address of this article: https://kexue.fm/archives/1904
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