A problem on James Stewart's book Essential Calculus with Differential Equations Section 2.6 Number 41 states that:
"Find all points on the curve
\begin{equation}
x^2y^2+xy=2
\end{equation}
where the slope of the tangent line is $-1$."
By reading the problem, I thought that the problem is easy and actually it is. Until I stumbled into something. Below is my solution to the problem and my questions will be written after I wrote my solution.
$\textbf{Solution:}$
To find all points $(x,y)$ such that the above equation has tangent line whose slope is $-1$, we will just simply use implicit differentiation to find $y'$ and equate it to $-1$; since $\left.y'\right|_{(x,y)}$ is the slope of the tangent line to the curve of the given equation at point $(x,y)$.
Doing that gives the equation
\begin{equation}
y'=-\frac{2xy^2+y}{2x^2y+x}=-1.
\end{equation}
The equation above implies $2xy^2+y=2x^2y+x$.
After some simplifications and by factoring by grouping we will arrive at:
\begin{equation}
(y-x)(2xy+1)=0.
\end{equation}
This means that $x-y=0$ that is $x=y$ or $2xy+1=0$. Knowing that $x=y$, we arrived at $2x^2+1=0$ whose roots are both imaginary.
At this point, I concluded that there is no such point $(x,y)$ such that $x^2y^2+xy=2$ has a tangent line with slope $-1$. But I stumbled for a while and think that it is irrational for the question to be included as an exercise in the book if it has no solution.
Where did I go wrong? Why am I wrong if i stopped at this part? Did I missed something?
Did I missed the other implication of $x=y$? That is; If $x=y$ then the original equation becomes:
\begin{equation}
x^2x^2+xx=2
\end{equation}
From here we get $x=\pm1$ and from here it follows that $y=1$ if $x=1$ and $y=-1$ if $x=-1$. Which means, the points at which the graph of the equation $x^2y^2+xy=2$ has a tangent lines with slope $-1$ are $(-1,-1)$ and $(1,1)$.
Is this answer correct?
Best Answer
The problem has solutions. Actually, all solutions satisfy the condition $y=x$. In fact if, in the equation $x^2y^2+xy=2$, you replace $y$ with $x$, you get $x^4+x^2=2$, which has two (and only two) real solutions: $\pm1$. So $\pm(1,1)$ are solutions of your problem. See the image below.