## Calculus 8th Edition

The equation of the tangent line is $$y=\frac{3}{4} x-\frac{1}{2}$$
$$x^{2}-x y-y^{2}=1, \ \ \ \ \ (2,1)$$ Since $x^{2}-x y-y^{2}=1$ by differentiate both sides by $x$, we get \begin{align} \Rightarrow &2 x-\left(x y^{\prime}+y \cdot 1\right)-2 y y^{\prime}=0\\ \Rightarrow &2 x-y=x y^{\prime}+2 y y^{\prime}\\ \Rightarrow& 2 x-y=(x+2 y) y^{\prime} \\ \Rightarrow& y^{\prime}=\frac{2 x-y}{x+2 y} \end{align} at $(x,y)=(2,1)$, we get $y^{\prime}=\frac{3}{4}.$ so an equation of the tangent line is $y-1=\frac{3}{4}(x-2)$ i.e $y=\frac{3}{4} x-\frac{1}{2}$