Semi axes of a shifted rectangular hyperbola

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

$$(x-c_0)(y-d_0)=A$$

In the classic case it is easy to find them:

$$(x/a)^2-(y/b)^2=1$$

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

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 $$a=b=\sqrt{2A}$$.

• Hello @BPP should they be segments? a and b are segments. – Gennaro Arguzzi May 5 at 16:50
• @GennaroArguzzi No they are lines just like the asymptotes $y=\pm\frac{b}{a}x$. – Paracosmiste May 5 at 17:02
• 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 – Gennaro Arguzzi May 5 at 19:47
• @GennaroArguzzi See the edit. – Paracosmiste May 5 at 20:05
• perfect explanation. – 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.