Prove that the curves xy=b and x²-y²=a intersect at right

Prove that the curves $xy=b$ and $x^{2}-y^{2}=a$ intersect at right angles for any values of $a$ and $b$.

Step 1: Slopes by implicit differentiation

\[ xy=b \;\Longrightarrow\; y+xy'=0 \;\Longrightarrow\; k_{1}=-\frac{y}{x}. \tag{1} \]

\[ x^{2}-y^{2}=a \;\Longrightarrow\; 2x-2yy'=0 \;\Longrightarrow\; k_{2}=\frac{x}{y}. \tag{2} \]

Step 2: Product of the slopes at a common point

At any point $(x_0,y_0)$ on both curves (necessarily $x_0\neq0$, $y_0\neq0$, since $x_0y_0=b\neq0$): \[k_{1}k_{2} = \left(-\frac{y_0}{x_0}\right)\!\left(\frac{x_0}{y_0}\right) = -1. \tag{3}\]