# Question #95baf

Mar 10, 2017

See below.

#### Explanation:

Mar 10, 2017

We have the points:

${P}_{1} = \left(- 1 , - 1\right)$ and ${P}_{2} = \left(1 , 1\right)$

#### Explanation:

The slope of the tangent line is given by $\frac{\mathrm{dy}}{\mathrm{dx}}$, which we can calculate by implicit differentiation:

$\frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{2} + x y\right) = 0$

$2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x {y}^{2} + y + x \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\left({x}^{2} y + x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - x {y}^{2} - y$

Since from the original equation we can see that $x y \ne 0$, divide both sides by $x y$

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

$\frac{x y + 1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x y + 1}{x}$

We can see that $x y \ne - 1$, otherwise, ${\left(x y\right)}^{2} + x y$ would be null, in contradiction with the original equation, so:

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

And we have:

$- \frac{y}{x} = - 1$

which means:

$x = y$

Substituting this in the original equation we find the values of $x$ for which $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$

$\left\{\begin{matrix}{x}^{2} {y}^{2} + x y = 2 \\ x = y\end{matrix}\right.$

${x}^{4} + {x}^{2} = 2$

${x}^{4} + {x}^{2} - 2 = 0$

Solving this as a second degree equation with unknown ${x}^{2}$:

${x}^{2} = \frac{- 1 \pm \sqrt{1 + 8}}{2}$

And choosing only the positive solution since ${x}^{2}$ cannot be negative for $x \in \mathbb{R}$:

${x}^{2} = 1$

$x = \pm 1$

Thus we have the points:

${P}_{1} = \left(- 1 , - 1\right)$ and ${P}_{2} = \left(1 , 1\right)$