I would like to know which are the semi axes of a shifted rectangular hyperbola:


In the classic case it is easy to find them:


$a$ and $b$ are the semi axes.

Thank you in advance.


The asymptotes are the lines $x=c_0$ and $y=d_0$. Of course $A\neq 0$.Since the asymptotes are perpendicular we have $a=b$.

The point $I(c_0,d_0)$ is the intersection of the asymptotes. The line $l: y=x-c_0+d_0$ passes through $I$ and is parallel to $y=x$. Assuming $A>0$ this line $l$ cut the hyperbola at its vertices. The intersection is obtained by solving $(x-c_0)^2=A$ which gives us the two vertices $R(c_0+\sqrt{A}, d_0+\sqrt{A})$ and $S(c_0-\sqrt{A}, d_0-\sqrt{A})$. The distance $ST$ is equal to $2a$ thus $a=b=\sqrt{2|A|}$ .

Another way to prove this: Let $x=\dfrac{u+v}{\sqrt{2}}+c_0$ and $y=\dfrac{-u+v}{\sqrt{2}}+d_0$. Then $(x-c_0)(y-d_0)=A$ gives us $v^2-u^2=2A$ which is the equation of an equilateral hyperbola and $a=b=\sqrt{2A}$.

  • $\begingroup$ Hello @BPP should they be segments? a and b are segments. $\endgroup$ – Gennaro Arguzzi May 5 at 16:50
  • $\begingroup$ @GennaroArguzzi No they are lines just like the asymptotes $y=\pm\frac{b}{a}x$. $\endgroup$ – Paracosmiste May 5 at 17:02
  • $\begingroup$ I think I did not formulated the question well. I am wondering which are the "equivalent" coefficient a and b in the case of a shifted rectangular hyperbola. They should exist and should be the sides of a rectangle. In the first figure here you can see the rectangle (but in this case the hyperbola is not shifted and is not rotate): encyclopediaofmath.org/index.php/Hyperbola $\endgroup$ – Gennaro Arguzzi May 5 at 19:47
  • $\begingroup$ @GennaroArguzzi See the edit. $\endgroup$ – Paracosmiste May 5 at 20:05
  • $\begingroup$ perfect explanation. $\endgroup$ – Gennaro Arguzzi May 6 at 10:39

Translation doesn’t affect semiaxis lengths or their orientations, so eliminate the irrelevant linear terms by translating the origin to $(c_0,d_0)$ to obtain the equation $x'y'=A$. Can you find the principal axes of this hyperbola? Once you’ve done that, translate back.


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