Find the condition for the curves $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $xy=c^{2}$ to intersect orthogonally.

(A) Let the curves intersect at $(x_{1}, y_{1})$.
For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,differentiating with respect to $x$ gives $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$,so $\frac{dy}{dx} = \frac{b^{2}x}{a^{2}y}$.
The slope of the tangent at $(x_{1}, y_{1})$ is $m_{1} = \frac{b^{2}x_{1}}{a^{2}y_{1}}$.
For the curve $xy=c^{2}$,differentiating with respect to $x$ gives $x\frac{dy}{dx}+y=0$,so $\frac{dy}{dx} = -\frac{y}{x}$.
The slope of the tangent at $(x_{1}, y_{1})$ is $m_{2} = -\frac{y_{1}}{x_{1}}$.
For orthogonal intersection,$m_{1} \times m_{2} = -1$.
Substituting the slopes,we get $\left(\frac{b^{2}x_{1}}{a^{2}y_{1}}\right) \times \left(-\frac{y_{1}}{x_{1}}\right) = -1$.
This simplifies to $-\frac{b^{2}}{a^{2}} = -1$,which implies $a^{2} = b^{2}$ or $a^{2}-b^{2}=0$.