Question

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

Answer 1

Differentiate both sides with respect to x and solve for the derivative dy/dx :

$\dfrac{\mathrm d}{\mathrm dx}\left[x^2y^2+xy\right] = \dfrac{\mathrm d}{\mathrm dx}[2] \\\\ \dfrac{\mathrm d}{\mathrm dx}\left[x^2\right]y^2 + x^2\dfrac{\mathrm d}{\mathrm dx}\left[y^2\right] + \dfrac{\mathrm d}{\mathrm dx}\left[x\right]y + x\dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ 2xy^2 + x^2(2y)\dfrac{\mathrm dy}{\mathrm dx} + y + x\dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ (2x^2y+x)\dfrac{\mathrm dy}{\mathrm dx} = -2xy^2-y \\\\ \dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{2xy^2+y}{2x^2y+x}$

This gives the slope of the tangent to the curve at the point (x, y).

If the slope of some tangent line is -1, then

$-\dfrac{2xy^2+y}{2x^2y+x} = -1 \\\\ \dfrac{2xy^2+y}{2x^2y+x} = 1 \\\\ 2xy^2+y = 2x^2y+x \\\\ 2xy^2-2x^2y + y - x = 0 \\\\ 2xy(y-x)+y-x = 0 \\\\ (2xy+1)(y-x) = 0$

Then either

$2xy+1 = 0\text{ or }y-x=0 \\\\ \implies y=-\dfrac1{2x} \text{ or }y=x$

In the first case, we'd have

$x^2\left(-\dfrac1{2x}\right)^2+x\left(-\dfrac1{2x}\right) = \dfrac14-\dfrac12 = -\dfrac14\neq2$

so this case is junk.

In the second case,

$x^2\times x^2+x\times x=x^4+x^2=2 \\\\ x^4+x^2-2 = (x^2-1)(x^2+2)=0$

which means either

$x^2-1 = 0 \text{ or }x^2+2 = 0 \\\\ x^2 = 1 \text{ or }x^2 = - 2$

The second case here leads to non-real solutions, so we ignore it. The other case leads to $x=\pm1$.

Find the y-coordinates of the points with x = ±1 :

$x=1 \implies y^2+y=2 \implies y=-2 \text{ or }y=1 \\\\ x=-1\implies y^2-y=2\implies y=-1\text{ or }y=2$

so the points of interest are (1, -2), (1, 1), (-1, -1), and (-1, 2).