How $y=\dfrac{1}{x}$ is derived from $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ hyperbola

Question

Unfortunately, I can't find the derivation of $y=\dfrac{1}{x}$ (1) from hyperbola canonical equation $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ (2) in the internet.

I tried a bit to derive it on my own, but unsuccessfully, and the reason why I posting it here is that I doubt, that it really can be derived only from given - without additional conditions, for example, assuming, that $a=b$, or something similar.

Basically, as I understand (1) is (2) with coordinate axis rotated by some angle (45 degrees?), but I do not know, how to express it mathematically. Perhaps, I need somehow change $x$ and $y$ in (2), and after that try to express to (1) form.

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Major update

I successfully derived the canonical equation (2), as a set of points, which absolute differenc of distances to two fixed points - focuses is constant.

In that case, I assumed, that focuses lie on X-axis. What if to just derive (1) the same way, assuming, that focuses are situated somewhere in first and third quarts?

Answer 1

Let's use $\left(-\frac{\pi}{4}\right)$-rotation matrix: $$\begin{bmatrix}\cos\frac{\pi}{4} & \sin\frac{\pi}{4} \\ -\sin\frac{\pi}{4} & \cos\frac{\pi}{4}\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix}\frac{x + y}{\sqrt 2} \\ \frac{-x+y}{\sqrt 2}\end{bmatrix}$$ Now we check how hyperbola form is tranformed: $$\left(\frac{x+y}{\sqrt 2}\right)^2 - \left(\frac{- x + y}{\sqrt 2}\right)^2 = 1$$ $$x^2 + 2xy +y^2 - \left(x^2 - 2xy + y^2\right) = 2$$ $$2xy = 1 \Rightarrow y = \frac{1}{2x}$$

Answer 2

Consider a plane with a hyperbola in it. Now impose $u$ and $v$ coordinate axes, in such a way that the hyperbola satisfies $v = \frac1u$. Now impose a new coordinate system in the same plane, with the same origin, with an $x$-axis going along the line $v = -u$ (positive direction in the fourth $uv$ quadrant) and a $y$-axis going along the line $v = u$ (positive direction in the first $uv$ quadrant). The units on the $xy$ axes are the same as they are on the $uv$ axes.

For any point with $uv$-coordinates $(u, v)$, the $xy$ coordinates of that same point are $\left(\frac1{\sqrt2}(u - v), \frac1{\sqrt2}(u + v)\right)$. On the other hand, if a point has $xy$-coordinates $(x, y)$, then the $uv$ coordinates of that point are $\left(\frac1{\sqrt2}(x + y), \frac1{\sqrt2}(x - y)\right)$. (If this isn't obvious to you, I strongly encourage you to actually take a pen and a sheet of paper, and draw these two coordinate systems on top of one another, and see that the coordinate translations are what I have said here.)

Now we turn our attention to the hyperbola. It consists of those points whose $uv$ coordinates $(u, v)$ satisfy the equation $uv = 1$ (which is to say, the product of the two coordinates is equal to $1$). But what if we have a point and it's $xy$ coordinates $(x, y)$ instead? How can we tell whether it is on the hyperbola? By using the above translation, of course!

If the $xy$ coordinates are $(x, y)$ then the $uv$ coordinates are $\left(\frac1{\sqrt2}(x + y), \frac1{\sqrt2}(x - y)\right)$. And we know that the point lies on the hyperbola if the product of their $uv$ coordinates equals $1$. So we get $$ \frac1{\sqrt2}(x + y) \cdot \frac1{\sqrt2}(x - y) = 1\\ \frac1{2}(x^2 - y^2) = 1\\ \frac{x^2}2 - \frac{y^2}2 = 1 $$ So this is the equation for that same hyperbola when using the $xy$ coordinate system.

Answer 3

You can put the equation of a hyperbola in the form $xy=k$ only if it is a rectangular hyperbola, i.e. if its asymptotes are perpendicular between them. That happens if transverse and conjugate axis are equal: $a=b$.

If you take as coordinate axes the axes of the hyperbola, you get then its equation as: $$ {x^2\over a^2}-{y^2\over a^2}=1. $$ The foci, in this case, lie at $(\pm\sqrt2a,0)$ and the above equation is that of the locus of points whose distances from the foci have a fixed difference $2a$.

To obtain the other equation you must take instead as coordinate axes the asymptotes of the hyperbola. Doing so, the foci of the same hyperbola as above have coordinates $\pm(a,a)$. You can check that the locus of points $(x,y)$ whose distances from the foci have a fixed difference $2a$ obeys in this case the equation $$ xy={a^2\over2}. $$

Answer 4

I also provide a solution, that combines my asumption, I wrote in the "major update" section of my post, supported by @EgorIvanov and @Intelligentipauca answers:

Since, hyperbola is a set of points, absolute diffrence of distances from each, to two fixed points, called focuses is constant, and equal some value $2a$, where $a$ - a constant, called major semi axis, the equation of hyperbola is:

$|\sqrt{(x-\phi_{1x})^2 + (y-\phi_{1y})^2}-\sqrt{(x-\phi_{2x})^2 + (y-\phi_{2y})^2}|=2a$ $(1)$

Where $\phi_1$ and $\phi_2$ are focuses points.

Hyperbola canonical equation $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is derived from $(1)$, considering focuses liying on X-axis and have coordinates $\phi_1(-c,0)$ and $\phi_2(c,0)$, where $c=\sqrt{a^2 + b^2}$, where $b$ - a constant, called minor semi axis.

1. To derive $y=\dfrac{1}{x}$ $(2)$ equation, as @Intelligentipauca mentioned we need to restrict the hyperbola as rectengular hyberbola, since $(2)$ is exactly such hyperbola. To do that, we only need to consider $c=\sqrt{a^2 + a^2}=a\sqrt{2}$. Now focuses are $\phi_1(-a\sqrt{2},0)$ and $\phi_2(a\sqrt{2},0)$
2. Next, since, $(2)$ is actually rotated (and also scaled) on 45 degrees, relatively to $(1)$ hyperbola, using rotation transformation, @EgorIvanov mentioned, we need to rotate focuses by 45 degrees counterclockwise, so we will have focuses $\phi_1(-a,-a)$ and $\phi_2(a,a)$

Now, having initial conditions, that satisfy our needs, we can start to disassemble $(1)$:

$$|\sqrt{(x+a)^2 + (y+a)^2}-\sqrt{(x-a)^2 + (y-a)^2}|=2a$$

$$\sqrt{(x+a)^2 + (y+a)^2}=\sqrt{(x-a)^2 + (y-a)^2} ±2a$$

$$x^2+2xa+a^2+y^2+2ya+a^2=x^2-2xa+a^2+y^2-2ya+a^2±4a\sqrt{(x-a)^2 + (y-a)^2}+4a^2$$

$$4xa+4ya-4a^2=±4a\sqrt{(x-a)^2 + (y-a)^2}$$

$$x+y-a=±\sqrt{(x-a)^2 + (y-a)^2}$$

$$(x+y)^2-2a(x+y)+a^2=x^2-2xa+a^2+y^2-2ya+a^2$$

$$x^2+2xy+y^2-2ax-2ay+a^2=x^2-2xa+a^2+y^2-2ya+a^2$$

$$2xy=a^2$$

$$y = \dfrac{a^2}{2x}$$

Note, that as @EgorIvanov mentioned $(2)$ is also 2 times scaled hyperbola, not only rotated, so if we want to get exactly $(2)$, besides the rotation of focuses we also needed to divide $x$ by $2$ in $(1)$, but you can do it on your own.