Tangent plane proof
Consider the general ellipsoid
\[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{z{}^2}}{{{c^2}}} = 1\]At the point \( \left( {x_0},{y_0},{z_0} \right)\) the tangent plane can be written as
\[\frac{{x{x_0}}}{{{a^2}}} + \frac{{y{y_0}}}{{{b^2}}} + \frac{{z{z_0}}}{{{c^2}}} = 1\]
Proof
Remember that the normal vector to a surface is proportional to the gradient vector at that point. Also remember the equation of a plane written in vector notation is
\[{\bf{r}} \cdot {\bf{n}} = d\]
where r is the general position vector, n is the unit normal to the plane and d is the distance from origin to plane.
Now the gradient at point \( \left( {x_0},{y_0},{z_0} \right)\)is given by
\(\nabla F\left( {\begin{array}{*{20}{c}} {{x_0},}&{{y_0},}&{{z_0}} \end{array}} \right) = \)\(\left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{{a^2}}},}&{\frac{{2{y_0}}}{{{b^2}}},}&{\frac{{2{z_0}}}{{{c^2}}}} \end{array}} \right)\) = \(\lambda {\bf{n}}\)
for some value λ.
Now \({\bf{r}} \cdot {\bf{n}} = d\) can be written as
\(\left( {\begin{array}{*{20}{c}} {x,}&{y,}&z \end{array}} \right) \cdot \)\(\left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{\lambda {a^2}}},}&{\frac{{2{y_0}}}{{\lambda {b^2}}},}&{\frac{{2{z_0}}}{{\lambda {c^2}}}} \end{array}} \right)\) = \(\left( {\begin{array}{*{20}{c}} {{x_0},}&{{y_0},}&{{z_0}} \end{array}} \right) \cdot\)\( \left( {\begin{array}{*{20}{c}} {\frac{{2{x_0}}}{{\lambda {a^2}}},}&{\frac{{2{y_0}}}{{\lambda {b^2}}},}&{\frac{{2{z_0}}}{{\lambda {c^2}}}} \end{array}} \right)\)
and upon expanding we get
\[\frac{{2x{x_0}}}{{{a^2}}} + \frac{{2y{y_0}}}{{{b^2}}} + \frac{{2z{z_0}}}{{{c^2}}} = \frac{{2x_0^2}}{{{a^2}}} + \frac{{2y_0^2}}{{{b^2}}} + \frac{{2z_0^2}}{{{c^2}}}\]
and by cancelling the 2s and realising the RHS satisfies the ellipsoid equation we arrive at
\[\frac{{x{x_0}}}{{{a^2}}} + \frac{{y{y_0}}}{{{b^2}}} + \frac{{z{z_0}}}{{{c^2}}} = 1\]