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

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

$$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4) = \frac{d}{dx}(0)$$

Using the power rule, product rule for $xy$, and chain rule for $y^2$:

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

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

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

Factor out $\frac{dy}{dx}$:

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

Isolate $\frac{dy}{dx}$:

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

This can also be written as:

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

Substitute $x=2$ and $y=4$ into the expression for $\frac{dy}{dx}$:

$$m = \frac{2(2) + 4}{2(4) - 2}$$ $$m = \frac{4 + 4}{8 - 2}$$ $$m = \frac{8}{6}$$ $$m = \frac{4}{3}$$

Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (2, 4)$ and $m = \frac{4}{3}$:

$$y - 4 = \frac{4}{3}(x - 2)$$

Multiply both sides by 3 to eliminate the fraction:

$$3(y - 4) = 4(x - 2)$$ $$3y - 12 = 4x - 8$$

Rearrange the terms to the standard form $Ax + By + C = 0$:

$$0 = 4x - 3y - 8 + 12$$ $$4x - 3y + 4 = 0$$

The equation of the tangent to the curve at $k(2, 4)$ is: