tangent lines to x^2 y^2 + xy = 2 where slope is -1

[ku1005]

With the following problem, could someone please tell me what I am doing wrong, or why my undestanding is incorrect? Thank you!

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

Via implicit differentiation, I get:

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

so

. . .-1 = -(y+2xy^2) / (2x^2y +x)

. . .2x^2y + x = y + 2xy^2

. . .x(2xy +1) = y (2xy +1)

. . .x/y = 1

From here, I thought that any point lying on the line y = x will satisfy the equation. However, I know this isn't correct: Suppose you choose (x, y) = (4, 4). This point doesn't even lie on the original curve, therefore it can't be correct.

The answers are given as (-1, -1) and (1, 1), which I understand are solutions. But I don't understand why my thinking above doesnt work. Would someone be able to tell me why? Perhaps I have solved it wrongly...?

Thank you!

 

[arthur ohlsten]

I believe there are two points,[1,1] and [-1,-1]

let xy=z
the equation becomes z^2+z-2=0
find the roots
z=-1+/-[1+8]^1/2 all over 2
z=-1+/-3 all over 2
z=1 or -2

the solution to
x^2y^2+xy-2=0 is:
xy=1 or xy=-2

the slope is x/y=1
x=y

for xy=1 and x=y
x^2=1
x=+/-1
[x=1 y=1] and [x=-1 y=-1] 2 points answer

for xy=-2 and x=y
x^2=-2 no real roots answer

Arthur

 

[soroban]

Re: tangent lines of curves

Hello, ku1005!

Arthur is absolutely correct.
. . I saw a back-door approach . . .

Find all points on the curve \(\displaystyle x^2y^2\,+\,xy\:=\:2\)
where the slope of the tangent line is \(\displaystyle \,-1.\)

We have: \(\displaystyle \:x^2y^2\,+\,xy\:=\:2\;\;\Rightarrow\;\;x^2y^2\,+\,xy\,-\,2\:=\:0\)

. . which factors: \(\displaystyle \xy\,-\,1)(xy\,+\,2)\:=\:0\)

And we have: \(\displaystyle \:\begin{Bmatrix}xy\:=\:1 \\ xy\:=\:-2\end{Bmatrix}\) . . . a pair of hyperbolas

The graph looks like this:

                         *|
                          |
                        * |*
                       *  |
                     *    | *
                  *       |  *
              *           |    *
        *                 |         *
      - - - - - - - - - - + - - - - - - - - - -
                *         |                 *
                     *    |           *
                       *  |       *
                        * |    *
                          |  *
                         *| *
                          |
                          |*
                          |

A slope of -1 occurs in Quadrants I and III only.
Hence, we are concerned with the branch: \(\displaystyle \,xy\:=\;1\)

We have: \(\displaystyle \:y\:=\:x^{-1}\;\;\Rightarrow\;\;y'\:=\:-x^{-2}\)

The slope is -1 when: \(\displaystyle \:-x^{-2} \:=\:-1\;\;\Rightarrow\;\;x\:=\:\pm1\)


Therefore, the points are: \(\displaystyle \1,\,1)\) and \(\displaystyle (-1,\,-1)\)

 

[ku1005]

thanks guys...didnt think of that