How to get the equation of the tangent surface of $F=x^2 + 2y^2 +3 z^2=21$, which is also parallel to the plane surface $ x+ 4y + 6z=0 $?
Here is what I've tried:
$n = \{1, 4, 6\}$ from $ x+ 4y + 6z=0 $ could be considered as the normal vector for the surface $F$.
$$ F_x = 2x, F_y=4y, F_z = 6z, $$
let $2x = 1, 4y = 4, 6z = 6$, then the tangent point is $(\frac12, 1, 1)$.
the equation representing the tangent surface is $ (x- \frac12) + 4(y-1) + 6(z-1) = 0$.
But I know the result is wrong. Can anyone help on this?
It is true that the surface you've found is parallel to the plane $x + 4y + 6z = 0$, and it is tangent to a level curve of $F$, but the trouble is that it isn't tangent to the particular level curve $F = 21$.
We need a point on $F = 21$ with gradient parallel to the one you've calculated, which amounts to solving \begin{eqnarray*} 2x = \lambda \\ 4y = 4\lambda \\ 6z = 6\lambda \\ x^2 + 2y^2 + 3z^2 = 21 \end{eqnarray*}
where $\lambda \neq 0$. Substituting the first three equations into the last, you should be able to find $\lambda$. This gets you your $(x,y,z)$ coordinate, and you know what to do from there.