I have to find the points such that the tangent plane of the surface $x^2+3y^2+4z^2-2xy=16$ are parallel to the plane $xz$.
So I compute $\nabla f(x,y,z)= (2x-2y,6y-2x, 8z)$ and therefore $\nabla f(x,y,z)=(0,1,0)$ where $(0,1,0)$ is the normal vector of the plane $xz$ but I have this equation's system.
\begin{eqnarray}
\left.
\begin{array}{rcl}
2x-2y & = & 0
\\ 6y-2x & = & 1
\\ 8z & = & 0
\end{array}
\right\}
\end{eqnarray}
and I conclude the solution and the point that I found is $(\frac{1}{4}, \frac{1}{4}, 0)$ and this point, give the the plane $y=\frac{1}{4}$ but this point isn't on the surface and this plane isn't tangent to the surface. What's wrong? I'll be grateful if you can give some advice with this.
Best Answer
In order for the tangent plane to be parallel to $xz$, $\nabla f(x,y,z)$ must be collinear to $(0,1,0)$, i.e. for some $k\in\mathbb R$ \begin{eqnarray} \left\{ \begin{array}{rcl} 2x-2y & = & 0\cdot k \\ 6x-2y & = & 1\cdot k \\ 8z & = & 0 \cdot k. \end{array} \right. \end{eqnarray} Notice that if $k = 0$, then we obtain a singular point at which the tangent plane may not exist.
In addition, the point of tangency must belong to the surface, that is, the equation $$x^2+3y^2+4z^2-2xy=16$$ must be satisfied. Thus, we obtain the system of equations \begin{eqnarray} \left\{ \begin{array}{rcl} 2x-2y & = & 0 \\ 6x-2y & = & k \\ 8z & = & 0 \\ x^2+3y^2+4z^2-2xy&=&16. \end{array} \right. \end{eqnarray} From the first three equations $x=y$, $z=0$, $k=4x$. Substituting this into the equation of the surface, we obtain $$ x^2+3x^2-2x^2=16 $$ $$ 2x^2=16 $$ $$ x^2=8. $$ Hence there are two points where the tangent plane is parallel to the plane $xz$: $(2\sqrt2,2\sqrt2,0)$ and $(-2\sqrt2,-2\sqrt2,0)$. Notice that $k=\pm8\sqrt2 \ne 0$.