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.
Thanks in advance.

  • $\begingroup$ Where are the foci of $xy=c^2$? $\endgroup$ – metamorphy Aug 28 '18 at 15:09
  • $\begingroup$ @metamorphy- One each on $x=\pm y$. $\endgroup$ – 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.
$$\Rightarrow (x^2-y^2)\sin\theta+2xy\cos\theta=2c^2$$


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