Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 51

Determine all points on the curve $x^2y^2 + xy = 2$ where the slope of the tangent line is $-1$.

Taking the derivative of the equation implicitly we have,

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


$\begin{equation} \begin{aligned} & \frac{d}{dx} (x^2) \cdot y^2 + x^2 \cdot \frac{d}{dy} (y^2) \frac{dy}{dx} + \frac{d}{dx} (x) \cdot y + x \cdot \frac{d}{dy} (y) \frac{dy}{dx} = 0 \\ \\ & 2xy^2 + 2x^2 y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \\ \\ & \frac{dy}{dx} = \frac{-2xy^2 - y}{2x^2 y + x} \end{aligned} \end{equation}$


But we are looking for the target lines that has slope $-1$, so

$\displaystyle \frac{dy}{dx} = -1$



$\begin{equation} \begin{aligned} & -1 = \frac{-2xy - y}{2x^2 y + x} \longrightarrow \cancel{-}1 = \cancel{-} \left( \frac{2xy^2 + y}{2x^2 y + x} \right) \\ \\ & 2x^2y + x = 2xy^2 + y \\ \\ & (y-x)(2xy + 1) = 0 \end{aligned} \end{equation}$




So,

$y = x$ and $2xy = -1$ or $\displaystyle xy = \frac{-1}{2}$

Substituting $\displaystyle xy = \frac{-1}{2}$ in the equation of the curve.


$\begin{equation} \begin{aligned} & x^2y^2 + xy = 2 \\ \\ & \left( \frac{-1}{2} \right) ^2 + \left( \frac{-1}{2} \right) = 2 \\ \\ & \frac{1}{4} - \frac{1}{2} = 2 \end{aligned} \end{equation}$


So there are no symbols. Next, we substitute $y = x$ into the equation of the curve we have..


$\begin{equation} \begin{aligned} x^2y^2 + xy =& 2 \\ \\ x^2(x^2) + x(x) =& 2 \\ \\ x^4 +x^2 =& 2 \\ \\ x^4 +x^2 - 2 =& 0 \end{aligned} \end{equation}$


By factoring, we have

$(x^2 - 1)(x^2 + 2) = 0$

The real solution are then..

$x = \pm 1$ but $y = x$ so $y = \pm 1$

Therefore, the points on the curve where the slope of the tangent line is $-1$ are $(1,1)$ and $(-1, -1)$.