Question

Find the points at which slope of the tangent on the curve `x^2y^2+xy=2` is `-1`?

Answer 1

Here is an outline. I've omitted details.

`x^2y^2 + xy = 2`

`dy/dx = -(2xy^2+y)/(2x^2y+x)`

Set `dy/dx = -1` and then use algebra to get

`2xy^2-2x^2y+y-x=0`.

Which is equivalent to

`2xy(y-x)+(y-x)=` and consequently to

`(2xy+1)(y-x) = 0`

So either

`2xy+1 = 0` and `y=-1/(2x)`

or `y=x`

Replacing `y` by each of these in the original `x^2y^2 + xy = 2`, we get

no solution when `y=-1/(2x)`

and

`x=+-1` when `y=x`.

Answer 2

See below.

Explanation:

`x^2y^2+xy-2=0` or `(xy)^2+(xy)-2=0`

solving for `(xy)`

`(xy)=(-1pm sqrt(1+8))/2 = {(-2),(1):}`

so we have

`x^2y^2+xy-2=0->{(xy=-2),(xy=1):}`

The slopes of `-1` occurs in

`xy = 1` at points `(-1,-1)` and `(1,1)` because for `xy=1`

`(dy)/(dx)=-y/x=-1/x^2=-1->x=pm1`

Answer 3

At points `(1,1)` and `-1.-1)`

Explanation:

Slope of a tangent at a point on the curve `x^2y^2+xy=2` will be given by the value of `(dy)/(dx)` at that point. So let us find its derivative, which is given by,

`2x xxy^2+x^2xx2yxx(dy)/(dx)+1xxy+x xx(dy)/(dx)=0`

i.e. `(dy)/(dx)[2x^2y+x]=-2xy^2-y` and

`(dy)/(dx)=-(2xy^2+y)/(2x^2y+x)`

= `-(y(2xy+1))/(x(2xy+1))=-y/x`

and as we have to identify, where slope of tangent is `-1`, we have solution `y=x`

and as `x^2y^2+xy=2`, this is at

`x^4+x^2-2=0`

or `(x^2-1)(x^2+2)=0`

or `x^2-1=0`

i.e. `x=+-1` and as `y=x`

we have slope of tangent as `-1` at `(1,1)` and `(-1,-1)`
graph{(x^2y^2+xy-2)(x-y)(y+x-2)(y+x+2)=0 [-10, 10, -5, 5]}