Differentiate the equation implicitly with respect to x to find dy/dx.

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Q1) Find the equation of the tangent to the curve $x^2 - xy - y^2 + 4 = 0$ at $K(2, 4)$ .

Differentiate the equation implicitly with respect to $x$ to find $\frac{dy}{dx}$ .
$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4) = \frac{d}{dx}(0)$ $2x - \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 2y \frac{dy}{dx} + 0 = 0$ $2x - y - x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$
Rearrange the equation to solve for $\frac{dy}{dx}$ .
$2x - y = x\frac{dy}{dx} + 2y\frac{dy}{dx}$ $2x - y = (x + 2y)\frac{dy}{dx}$ $\frac{dy}{dx} = \frac{2x - y}{x + 2y}$
Evaluate the slope of the tangent at the point $K(2, 4)$ by substituting $x=2$ and $y=4$ .
$m = \frac{2(2) - 4}{2 + 2(4)}$ $m = \frac{4 - 4}{2 + 8}$ $m = \frac{0}{10}$ $m = 0$
Use 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 $\boxed{y=4}$ .

Differentiate the equation implicitly with respect to x to find (dy)/(dx).