Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 50

Determine the equation of the tangent line to the curve $\displaystyle y = x^2+4xy+y^2=13$ at the point $(2,1)$

Solving for the slope of the tangent line $(m_T)$, where $y'=m_T$


$\begin{equation} \begin{aligned} \frac{d}{dx} (x^2)+ 4 \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) &= \frac{d}{dx} (13)\\ \\ \frac{d}{dx} (x^2) + 4 \left[(x) \frac{d}{dx}(y) + (y) \frac{d}{dx}(x) \right] + \frac{d}{dx} (y^2) &= \frac{d}{dx} (13)\\ \\ 2x + 4 \left[ (x) \frac{dy}{dx} + (y)(1) \right] + 2y \frac{dy}{dx} &= 0\\ \\ 2x + 4x \frac{dy}{dx} + 4y + 2y \frac{dy}{dx} &= 0\\ \\ 2x + 4xy' + 4y + 2yy' &= 0\\ \\ 4xy' + 2yy' &= -2x-4y\\ \\ y'(4x+2y) &= -2x-4y\\ \\ \frac{y'\cancel{(4x+2y)}}{\cancel{4x+2y}} &= \frac{-2x-4y}{4x+2y}\\ \\ y' &= \frac{-2x-4y}{4x+2y} \end{aligned} \end{equation}$


For $x = 2$ and $y = 1$, we obtain


$\begin{equation} \begin{aligned} y' = m_T &= \frac{-2(2)-4(1)}{4(2)+2(1)}\\ \\ m_T &= \frac{-4-4}{8+2}\\ \\ m_T &= \frac{-8}{10}\\ \\ m_T &= \frac{-4}{5} \end{aligned} \end{equation}$


Using point slope form


$\begin{equation} \begin{aligned} y - y_1 &= m_T (x - x_1)\\ \\ y - 1 &= \frac{-4}{5}(x-2)\\ \\ y - 1 &= \frac{-4x+8}{5}\\ \\ y &= \frac{-4x+8}{5} +1 \\ \\ y &= \frac{-4x+8+5}{5}\\ \\ y &= \frac{-4x + 13}{5} \qquad \text{or} \qquad y = \frac{13-4x}{5} && \Longleftarrow \text{(Equation of the tangent line at (2,1))} \end{aligned} \end{equation}$