# Find all points on the curve x^2y^2+xy=2 where the slope of the tangent line is -1?

Mar 11, 2015

Use Implicit Differentiation (and hope it all falls out the bottom)

Differentiate both sides of the equation:
$D \left({x}^{2} {y}^{2} + x y\right) = D \left(2\right)$

using the product rule (a couple of times) and the chain rule:
$\left(2 x \cdot {y}^{2} + {x}^{2} \cdot 2 y \cdot y '\right) + \left(\left(1\right) \cdot y + x y '\right) = 0$

giving
$y ' \left(2 y {x}^{2} + x\right) + 2 x {y}^{2} + y = 0$
or
$y ' = - \frac{2 x {y}^{2} + y}{2 y {x}^{2} + x}$

We are asked for the points where the slope ($y '$) equals $- 1$
so
$- \frac{2 x {y}^{2} + y}{2 y {x}^{2} + x} = - 1$
or
$2 x {y}^{2} + y = 2 y {x}^{2} + x$

$y \left(2 x y + 1\right) = x \left(2 x y + 1\right)$

provided $x y \ne - \frac{1}{2}$ (which can be ruled out as extraneous by referring to the original equation
we have
$y = x$

The original equation becomes
${x}^{2} \cdot {x}^{2} + x \cdot x = 2$

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

$\left({x}^{2} + 2\right) \cdot \left({x}^{2} - 1\right) = 0$

since ${x}^{2} + 2 = 0$ is obviously extraneous
${x}^{2} = \pm 1$

with (as already determined $y = x$)
the required points are
$\left(- 1 , - 1\right)$ and $\left(1 , 1\right)$