Question

Prove that $xy=c^2$ is other form of rectangular hyperbola.

Answer 1

the equation of rectangular hyperbola is $x^2-y^2=a^2$

$(x+y(x-y)=a^2$

we know that $x=y=0$ and $x-y=0$

are at ${45}^o$ and ${135}^o$ to the x-axis. Now if we can rotate the axes through $\theta ={45}^o$ without changing the origin so we can replace $(x,y)$ by $(x\cos \theta -y\sin \theta ,x\sin \theta +t\cos \theta )$

i.e., $(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}})$

The equation $x^2-y^2=a^3$ becomes

${\left(\frac{x+y}{\sqrt{2}}\right)}^2-{\left(\frac{x+y}{\sqrt{2}}\right)}^2=a^2$

$\Rightarrow$ $\frac{4xy}{2}=a^2$

$\Rightarrow$ $xy=\frac{a^2}{2}$

$xy=c^2$ ($\because \frac{a^2}{2}=c^2$)

Therefore $xy=c^2$ is the another form of rectangular hyperbola