Solution.

Let \(\displaystyle {F(x,y,z) = x^2 + \frac {y^2}{4} + \frac {z^2}{9} - 3 = 0}\), from this we get \(\displaystyle {\nabla F(x,y,z) = 2x\textbf {i} + \dfrac {1}{2}y\textbf {j} + \dfrac {2}{9}z\textbf {k}}\).

\(\displaystyle {\nabla F(1, 2, 3) = 2\textbf {i} + \textbf {j} + \dfrac {2}{3}\textbf {k}}\), which is normal vector to the plane through at point \((1,2,3)\). Thus, equation of the plane \[2(x -1) + (y - 2) + \dfrac {2}{3}(z - 3) = 0 \hspace {0.4cm}\text {OR}\hspace {0.4cm} 6x + 3y + 2z = 18\]

Now suppose \(f\) has continuous partial derivatives an equation of the tangent plane to the surface \(z = f(x,y)\) at the point \(P(x_0,y_0,z_0)\) is \[z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0)\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (I)\]

Solution.

Let \(f(x,y) = 2x^2 + y^2\hspace {0.2cm} , \hspace {0.2cm} f_x = 4x\) and \(f_y = 2y\). Then \(f_x(1,1) 4\hspace {0.2cm} , \hspace {0.2cm} f_y (1,1) =2\), so that \begin {align*} z - z_0 & = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0)\\ z - 3 & = 4(x - 1) + 2(y - 1)\\ z & = 4x + 2y - 3\\\\ \end {align*}