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Prove that 1/x is a hyperbola with foci given by $(\sqrt2,\sqrt2)$ and $(-\sqrt2,-\sqrt2)$.

The idea is to use ONLY the definition of a hyperbola, that is:

Let $F_1$ and $F_2$ be two given points. A hyperbola $H$ is the set of points in the plane such that $$|dist(P,F_1)-dist(P,F_2)| =|dist(Q,F_1)-dist(Q,F_2)|,$$ for all $P,Q$ in $H$.

To be more precise, I don't want to use any notion of matrices since this is having no background on linear algebra.

What I did was to pick a generic point of the form $(x,1/x)$ and work with $$|dist(F_1,P)-dist(F_2,P)|,$$ where $F_1=(\sqrt2,\sqrt2), F_2=(-\sqrt2,-\sqrt2)$ and $P=(x,1/x)$. The problem was that I cannot prove that the result is independent of $x$.

Since that didn't work and knowing that $(1,1)$ belongs to the hyperbola, I calculate the difference of the distances between that point and the foci and the result is

$$dist(F_2,(1,1))-dist(F_1,(1,1))=\sqrt2(\sqrt2+1)- \sqrt2(\sqrt2-1)=2\sqrt2.$$

Therefore, the constant value that I am expecting to get is $2\sqrt2$, which I can't get in the general case!

After that, I tried with $(2,1/2)$, the result was

$$dist(F_2,(2,1/2))-dist(F_1,(2,1/2))=\frac{\sqrt{33+20\sqrt2}}{2}-\frac{\sqrt{33-20\sqrt2}}{2} $$

The key is to notice that $$\frac{\sqrt{33\pm 20\sqrt2}}{2}= \frac{5}{2}\pm \sqrt2.$$ (that is according to Wolfram). However, I don't know how to make an algebraic arrangement in order to get that. With that in mind, I am hoping to get for the general case something like $$f(x)+\sqrt2-(f(x)-\sqrt2),$$ where of course $f(x)$ should be some number depending on $x$. However, I am not being able to do it.

Any help is appreciated.

Thanks.

Math Guy
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  • What do you mean by matrices? You don't want to rotate axes? And $1/x$ does not define a curve. It's not even an equation. Perhaps you wanted to write something like $xy=1$? – Allawonder May 24 '19 at 23:24
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    @Allawonder The graph is $y=1/x$ from context. – Compact May 24 '19 at 23:26
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    ( the matrix solution that is not wanted is here - https://math.stackexchange.com/questions/2840823/prove-that-the-graphic-of-frac-1-x-is-a-branch-of-hyperbola?rq=1 ) – Calvin Khor May 25 '19 at 22:35

1 Answers1

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Note that since $F_i$ are points, $$ dist(F_i,P) = |F_i - P|.$$ Setting $F = F_1=(\sqrt2,\sqrt2)$, we have $-F=F_2$. Then we are asked to check that

$$ \Big||P-F| - |P+F|\Big| \text{ is constant in $P$,}$$ where $P=(x,1/x)$ is an arbitrary point in the graph of $1/x$. We compute directly,

$$ |P\pm F|^2 = (x\pm \sqrt 2)^2 + (1/x \pm \sqrt2)^2 = x^2 + x^{-2} +4 \pm 2\sqrt{2}(x+x^{-1}) $$ The following is perhaps a trick, $$ x^2 + x^{-2} + 2 = (x+x^{-1})^2$$ Recognizing the importance of $x+x^{-1}$, let $t = x+x^{-1}$. Note that for $x\in\mathbb R\setminus \{0\}$, $|t|\ge \color{orange}{2}$. Then $$ |P\pm F|^2 = t^2 \pm 2\sqrt 2 t + (\sqrt{2})^2 = (t \pm \sqrt 2)^2$$ Therefore $$ |P+ F| - |P- F| = |t + \sqrt 2| - |t - \sqrt 2| = \begin{cases} 2\sqrt{2}&t>\color{blue}{\sqrt{2}},\\[0.6em] -2\sqrt{2} & t<-\color{blue}{\sqrt{2}},\\ \text{who cares} & \text{otherwise.} \end{cases}$$ Since $\color{blue}{\sqrt{2}}<\color{orange}2$, $\Big||P+ F| - |P- F|\Big| = 2\sqrt{2}$ for all $P$ of the form $(x,1/x)$, where $x\in \mathbb R\setminus \{0\} $.

Calvin Khor
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