Equilateral hyperbola

This hyperbola, in which $a=b$, is called equilateral. Hence the eccentricity is $e=\sqrt{2}$.

Multiplying by $a^2$ in the expression $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we get the equation $x^2-y^2= a^2$. In this case the asymptotes would be $y=x$, $y =-x$.

It is possible to observe that the asymptotes are orthonormals. It would then be interesting if they were to coincide with our orthonormal axes. To do so, all we need is a $45$ degree turn. The resultant equation $x \cdot y=\frac{a^2}{2}$ can be expressed as $\displaystyle y=\frac{k}{x}$ by generating the following diagram:

Another expression, in which the hyperbola will not be in the first quadrant anymore is $\displaystyle y=-\frac{k}{x}$, giving place to:

Consider the hyperbola $y=-\frac{8}{x}$, and find its eccentricity and its focal distance.

The eccentricity is, by definition, of an equilateral hyperbola $e=\sqrt{2}$.

To identify $\displaystyle k=8=\frac{a^2}{2}$, then $a= \sqrt{16}=4$.

As $a=b$, with $c^2=a^2+b^2$ is it $c= \sqrt{2 \cdot a^2}=a\sqrt{2}=4\sqrt{2}$.