# What is the general equation of a rectangular hyperbola whose axes pass through origin?

I know the general equation of rectangular hyperbola whose foci lie on x-axis which is $x^2-y^2=a^2$

But by changing values of $a$ we don't arrive to the general equation of hyperbola whose asymptotes are $x,y$ axes $xy=c^2$.

I don't know the equation of hyperbola whose axes pass through origin$(0,0)$.
I think rotation of coordinates will do but how? I'm confused when rotating hyperbola ${x^2\over a^2}- {y^2\over b^2}=1$ to get equation of hyperbola whose axes pass through origin and apply the theorem that length of conjugate and transverse axes are equal in rectangular hyperbola.

Any hint will do.
• Where are the foci of $xy=c^2$? – metamorphy Aug 28 '18 at 15:09
• @metamorphy- One each on $x=\pm y$. – Love Invariants Aug 28 '18 at 15:10
The hyperbola given by the equation $xy=c^2>0$ is your prime example of a rectangular hyperbola. Let's rotate it by some arbitrary angle $\phi$! This rotation maps points $(x,y)$ to new points $(x',y')$, whereby $$(x,y)^\top=\left[\matrix{\cos\phi&-\sin\phi\cr\sin\phi&\cos\phi\cr}\right](x',y')^\top\ .$$ This means that the coordinates of the preimage $(x,y)$ are obtained from the coordinates $(x',y')$ of the image by $$x=\cos\phi\> x'-\sin\phi\> y',\qquad y=\sin\phi\> x'+\cos\phi\> y'\ .$$ If $(x,y)$ is lying on the model hyperbola then $xy=c^2$. This implies that the coordinates of the image point $(x',y')$ have to satisfy $$(\cos\phi\> x'-\sin\phi\> y')(\sin\phi\> x'+\cos\phi\> y')=c^2\ .$$ This can be rewritten as $$\cos\phi\sin\phi(x'^2-y'^2)+(\cos^2\phi-\sin^2\phi)x'y'=c^2\ .$$ Leaving away the primes here gives the result of the question.
$$\Rightarrow (x^2-y^2)\sin\theta+2xy\cos\theta=2c^2$$