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I came to a question in James Stewart's "Calculus Early Transcendentals" about implicit differentiation saying that:

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

After implicit differentiation I came to $$y'x(2xy+1)+y(2xy+1)=0\tag{1}$$ $$(2xy+1)(y'x+y)=0$$ Then $xy=\frac{-1}{2}$ or $y'=\frac{-y}{x}$

But by pluging $xy=\frac{-1}{2}$ into the original equations we get $$x^2y^2+xy=\frac{1}{4}-\frac{1}{2}\ne2$$ So, we conclude that $xy=\frac{-1}{2}$ is an extraneous solution. Then what caused it to be so ?? Q1

Also, we could start from equation $(1)$ to reach that $$y'=\frac{-y(2xy+1)}{x(2xy+1)} \tag{2}$$ We need the tangent to be $-1$. So we equate $y'=-1$ $$\frac{y(2xy+1)}{x(2xy+1)}=1$$ We now multiply by $x(2xy+1)$ to get $$y(2xy+1)=x(2xy+1)$$ I am now afraid to divide by $2xy+1$ in order not to get a missing solution. [Is it wise not to divide ?? Q2 When to remove the common factor of the numerator and denominator and not afraid of missing solutions ?? Q3]

So, I subtracted RHS-LHS to get $$y(2xy+1)-x(2xy+1)=0$$ Factorizing by grouping $$(y-x)(2xy+1)=0$$

Now we reached our result $y=x$. Sadly along with the extraneous solution. Is multiplying by the denominator of the rational equation $(2)$ is the cause of the second appearance of the extraneous solution ?? Q4

Moo
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Zyad Yasser
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    Solve for $xy$ may be a method. – jonsno Apr 26 '18 at 12:14
  • I don't understand your question! You have yourself verified that $2xy+1$ is not satisfied by our original curve, so indeed we have $y'=-y/x$ and subsequently $y=x$. Put $y=x$ in original equation to get points of tangency.. – jonsno Apr 26 '18 at 12:32

3 Answers3

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After implicit differentiation I came to $$y'x(2xy+1)+y(2xy+1)=0\tag{1}$$ $$(2xy+1)(y'x+y)=0$$

This is correct. Now rather than solving this for $y'$, you could simply substitute $y'=-1$, since this is what you want, and add the equation of the curve to get the following system: $$\left\{\begin{array}{rcl} \left(\color{red}{2xy+1}\right)\left(\color{blue}{y-x}\right) & = & 0 \\ (xy)^2+xy & = & 2 \end{array}\right.$$ You need to consider both equations since you're looking for points:

  • on the curve (i.e. satisfying the second equation);
  • and with the correct slope (i.e. satisfying the first equation).

Now from the first equation you have either $\color{red}{xy=-\tfrac{1}{2}}$ or $\color{blue}{y=x}$. You say this first case corresponds to an "extraneous solution" but it's not a solution since no points lying on the curve satisfy this equation. Solutions are only those points satisfying both equations.

Substitution of $y=x$ into the second equation leads to $x=\pm 1$ which then leads to the two correct solutions, namely the points $(1,1)$ and $(-1,-1)$.

StackTD
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$$ x^2y^2+xy-2 = 0 $$

Calling $z = xy \Rightarrow z^2+z-2 = (z+2)(z-1) = (xy+2)(xy-1) = 0 \Rightarrow\left\{ \begin{array}{rcl} xy +2 & = & 0\\ xy - 1 & = & 0 \end{array}\right.$

Attached a plot of $(xy+2)(xy-1) = 0$ In red $xy+2=0$ and in blue $xy-1 = 0$

enter image description here

Hence the tangency should be determined over $xy-1=0$ with two solutions for $x = \pm 1$

Cesareo
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You are over thinking it.

Start with

$$ (2xy+1)(y'x+y)=0 $$

Set $y'= -1$:

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

So $(2xy+1)=0$ and/or $(-x+y)=0$.

$x=y$ is always a solution whether $(2xy+1)=0$ or not. Since x,y are real, the and condition is never met.

If $(2xy+1)=0$ then $(y'x+y)$ can be any value and doesn't tell you anything about $y'$

Now plug $y=x$ into the original equation to solve for the points.

Hope this helps.

Ced

Cedron Dawg
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