Two curves in a drawing, xy = 4 and x² - y² = 15, are used to make a net of small curved squares…

Step 1. The point. $4 \times 1 = 4$ and $16 - 1 = 15$. Both equations hold.

Step 2. The first curve. $y + x\,\dfrac{dy}{dx} = 0$ gives $\dfrac{dy}{dx} = -\dfrac{y}{x} = -\dfrac{1}{4} = -0.25$.

Step 3. The second curve. $2x - 2y\,\dfrac{dy}{dx} = 0$ gives $\dfrac{dy}{dx} = \dfrac{x}{y} = 4$.

Step 4. The angle. $m_1m_2 = -0.25 \times 4 = -1$, so the tangents are perpendicular. The angle is $90^\circ$.

The second point. At $(-4, -1)$ the gradients are $-\dfrac{y}{x} = -\dfrac{-1}{-4} = -0.25$ and $\dfrac{x}{y} = \dfrac{-4}{-1} = 4$, so the angle is again $90^\circ$.

In general. The two gradients are $-\dfrac{y}{x}$ and $\dfrac{x}{y}$, whose product is $-1$ at every common point. The curves are orthogonal everywhere.