Find the equation of the tangent to the curve x2 − xy − y2 + 4 = 0 at the point K(2,4).

Substitute $x=2$ and $y=4$ into the equation of the curve $x^2 - xy - y^2 + 4 = 0$:

$$(2)^2 - (2)(4) - (4)^2 + 4$$ $$4 - 8 - 16 + 4$$ $$-4 - 16 + 4$$ $$-20 + 4 = -16$$

Since $-16 \neq 0$, the point $K(2,4)$ does not lie on the curve. However, as the question asks for the tangent at this point, we will proceed by finding the slope of the curve at $K(2,4)$ and using it to form the tangent line. This implies there might be a typo in the question, and we are treating $K(2,4)$ as the point of tangency.

The equation is $x^2 - xy - y^2 + 4 = 0$.
Differentiate each term:

$$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4) = \frac{d}{dx}(0)$$ $$2x - \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 2y \frac{dy}{dx} + 0 = 0$$ $$2x - y - x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$$

Group terms containing $\frac{dy}{dx}$:

$$2x - y = x\frac{dy}{dx} + 2y\frac{dy}{dx}$$ $$2x - y = (x + 2y)\frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{2x - y}{x + 2y}$$

Now, substitute the coordinates of point $K(2,4)$ ($x=2, y=4$) into the derivative to find the slope $m$:

$$m = \frac{2(2) - 4}{2 + 2(4)}$$ $$m = \frac{4 - 4}{2 + 8}$$ $$m = \frac{0}{10}$$ $$m = 0$$

Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (2,4)$ and $m=0$:

$$y - 4 = 0(x - 2)$$ $$y - 4 = 0$$ $$y = 4$$

The equation of the tangent to the curve at $K(2,4)$ is $\boxed{y=4}$.