## Related questions with answers

Find the points on the surface

$x y + y z + z x - x - z ^ { 2 } = 0$

where the tangent plane is parallel to the xy-plane.

Solution

VerifiedLet

$\begin{align} f(x,y,z)=xy+yz+zx-x-z^2=0 \end{align}$

Then the gradiend $\nabla f$ is

$\begin{align*} \nabla f&=f_x(x,y,z)\textbf{i}+f_y(x,y,z)\textbf{j}+f_z(x,y,z)\textbf{k}\\&= (y+z-1)\textbf{i}+(x+z)\textbf{j}+(y+x-2z)\textbf{k} \end{align*}$

Since we need to find the points when the given tangent plane is parallel to the $xy-$plane we have that the gradient $\nabla f$ is perpendicular to the $xy-$plane.

Hence we need to solve the system:

$\begin{align*} \begin{cases} y+z-1=0\\ x+z=0 \end{cases}\Rightarrow\begin{cases} y=1-z\\ x=-z \end{cases} \end{align*}$

Returning this into $(1)$ gives

$\begin{align*} (-z)(1-z)+(1-z)z+z(-z)-(-z)-z^2=0&\Rightarrow -z+z^2+z-z^2-z^2+z-z^2=0\\&\Rightarrow z-2z^2=\\&\Rightarrow z(1-2z)=0\\&\Rightarrow\color{#4257b2} z=0\vee z=\frac{1}{2} \end{align*}$

For $\color{#4257b2}z=0$ we get

$\begin{align*} \color{#4257b2}\begin{cases} y=1-0=1\\ x=0 \end{cases} \end{align*}$

and for $\color{#4257b2}z=\frac{1}{2}$ we get

$\begin{align*} \color{#4257b2}\begin{cases} y=1-\frac{1}{2}=\frac{1}{2}\\ x=-\frac{1}{2} \end{cases} \end{align*}$

Hence, the wanted points are

$\begin{align*} \color{#c34632}\boxed{(0,1,0),\left(-\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)} \end{align*}$

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