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 :

`(\mathrm d)/(\mathrm dx)\left[x^2y^2+xy\right] = (\mathrm d)/(\mathrm dx)[2] \\\\ (\mathrm d)/(\mathrm dx)\left[x^2\right]y^2 + x^2(\mathrm d)/(\mathrm dx)\left[y^2\right] + (\mathrm d)/(\mathrm dx)\left[x\right]y + x(\mathrm dy)/(\mathrm dx) = 0 \\\\ 2xy^2 + x^2(2y)(\mathrm dy)/(\mathrm dx) + y + x(\mathrm dy)/(\mathrm dx) = 0 \\\\ (2x^2y+x)(\mathrm dy)/(\mathrm dx) = -2xy^2-y \\\\ (\mathrm dy)/(\mathrm dx) = -(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

`-(2xy^2+y)/(2x^2y+x) = -1 \\\\ (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=-\frac1{2x} \text{ or }y=x`

In the first case, we'd have

`x^2\left(-\frac1{2x}\right)^2+x\left(-\frac1{2x}\right) = \frac14-\frac12 = -\frac14\\eq2`

so this case is junk.

In the second case,

`x^2* x^2+x* 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).