On which points of $xy=(1-x-y)^2$ is the tangent parallel to the $x$-axis?

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If the tangent line is parallel to the $x$-axis, then $y' = 0$.

We can determine when $y' = 0$ by differentiating the equation

$$xy = (1 - x - y)^2$$

implicitly with respect to $x$, which yields

\begin{align*} y + xy' & = 2(1 - x - y)(-1 - y')\\ y + xy' & = -2(1 - x - y) - 2(1 - x - y)y'\\ xy' + 2(1 - x - y)y' & = -2(1 - x - y) - y\\ (x + 2 - 2x - 2y)y' & = -2 + 2x + 2y - y\\ (2 - x - 2y)y' & = -2 + 2x + y\\ y' & = \frac{-2 + 2x + y}{2 - x - 2y} \end{align*} which is equal to $0$ when $y = 2 - 2x$.

Substituting $2 - 2x$ for $y$ in the original equation yields

\begin{align*} x(2 - 2x) & = [1 - x - (2 - 2x)]^2\\ 2x - 2x^2 & = (-1 + x)^2\\ 2x - 2x^2 & = 1 - 2x + x^2\\ 0 & = 3x^2 - 4x + 1\\ & = 3x^2 - 3x - x - 1\\ & = 3x(x - 1) - 1(x - 1)\\ & = (3x - 1)(x - 1) \end{align*}

Thus, $y' = 0$ when $x = 1$ or $x = 1/3$. If $x = 1$, then $y = 2 - 2x = 0$. If $x = 1/3$, then $y = 2 - 2x = 4/3$. Hence, the curve $xy = (1 - x - y)^2$ has horizontal tangents at the points $(1, 0)$ and $(1/3, 4/3)$.

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Updated on November 07, 2020

Comments

  • mao2047
    mao2047 about 3 years

    On which points of $xy=(1-x-y)^2$ is the tangent parallel to the $x$-axis?

    All I get is the derivative of the function, as far I know, I set the derivative equals to zero.

    • Rocket Man
      Rocket Man about 9 years
      That sounds about right.
    • mao2047
      mao2047 about 9 years
      But I don't know how to proceed
  • mao2047
    mao2047 about 9 years
    Yes! This is right, thank you!