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

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Solution 1

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.

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Solution 2

Note that in this case you can even solve for $y$. The equation is just a quadratic equation in $xy$ with solutions $xy = 1$ and $xy = -2$ so the curves are $y = 1/x$ and $y = -2/x$.

The derivatives then are $y' = -1/x^2$ and $y' = 2/x^2$. The first of these $y' = -1$ when $x = \pm 1$, while the second has no solution for $y' = -1.$

Solution 3

We may find the derivative $\dfrac{{\rm d}y}{{\rm d}x}$ as below

$$2xy^2+ x^2 \cdot 2y \cdot \frac{{\rm d}y}{{\rm d}x}+y+x \frac{{\rm d}y}{{\rm d}x}=0,$$ thus $$\frac{{\rm d}y}{{\rm d}x}=-\frac{2xy^2+y}{2x^2y+x}=-\frac{y(2xy+1)}{x(2xy+1)}=-\frac{y}{x}.$$

Let $$\frac{{\rm d}y}{{\rm d}x}=-\frac{y}{x}=-1.$$Hence, $$y=x.$$ Put it into the original equation. We obtain $$x^4+x^2-2=(x+1)(x-1)(x^2+2)=0.$$Solve it. We have $$x_1=-1,~~~x_2=1.$$ As a result, the points we want to find are $(-1,-1)$ and $(1,1)$.

Solution 4

The tangent line to the algebraic curve $f(x,y)=0$ at a simple point $(a,b)$ has equation $$ \frac{\partial f}{\partial x}(a,b)(x-a)+ \frac{\partial f}{\partial y}(a,b)(y-b)=0 $$ The slope is $-1$ if and only if $$ \frac{\partial f}{\partial x}(a,b)=\frac{\partial f}{\partial y}(a,b) $$ This leads to $$ \begin{cases} a^2b^2+ab=2 \\[4px] 2ab^2+b=2a^2b+a \end{cases} $$ The second equation simplifies to $(2ab+1)(a-b)=0$. So we either have $2ab=-1$ or $b=a$.

The case $2ab=-1$ is easily dismissed, so we remain with $b=a$ and the first equation gives $a^4+a^2-2=0$, so $a^2=1$ and $a=\pm1$.

The curve has no (proper) singular point, because at them both partial derivatives should vanish, but the previous system shows this can't happen.


The check for singular points should be done. Consider $x^3-x^2+y^2=0$. Your method would only catch one point, namely $(8/9,8/27)$, but the curve has a tangent with slope $-1$ also at the origin.


Actually the curve has singular points at infinity. Indeed, the homogeneous equation is $$ F(x,y,z)=x^2y^2+xyz^2-2z^4=0 $$ and $$ \frac{\partial F}{\partial x}=2xy^2+yz^2 \qquad \frac{\partial F}{\partial y}=2x^2y+xz^2 \qquad \frac{\partial F}{\partial z}=2xyz-8z^3 $$ The last derivative vanishes for $z=0$ or $xy=4z^2$. For $z=0$ we get either $x=0$ or $y=0$, leading to the improper points $(0,1,0)$ and $(1,0,0)$ (the improper points of the axes). For $xy=4z^2$ we get no solution.

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Updated on August 01, 2022

Comments

  • Jr Antalan
    Jr Antalan over 1 year

    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?

  • Jr Antalan
    Jr Antalan over 5 years
    I agree with you @egreg. However, the method that we are allowed to use is the one without partial derivatives. If we are only allowed to use partial derivatives. But your answer is very informative. Thank you.