Find an equation of the tangent plane to the given surface at the specified point.

\(\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},{\left({3},-{2},{18}\right)}\)

\(\displaystyle\Rightarrow{z}_{{x}}={4}{\left({x}-{1}\right)}\)

\(\displaystyle={4}{x}-{4}\),

\(\displaystyle{z}_{{y}}={12}{\left({y}+{3}\right)}\)

\(\displaystyle={12}{y}+{36}\)

At the point (3,-2,18):

\(\displaystyle{z}_{{x}}={4}{x}-{4}\)

\(\displaystyle={4}{\left({3}\right)}-{4}\)

\(\displaystyle={12}-{4}\)

\(\displaystyle={8}\)

\(\displaystyle{z}_{{y}}={12}{y}+{36}\)

\(\displaystyle=-{12}{\left(-{2}\right)}+{36}\)

\(\displaystyle=-{24}+{36}\)

\(\displaystyle={12}\)

Equation of the tangent plane is,

\(\displaystyle{z}-{z}_{{1}}={z}_{{x}}{\left({x}-{x}_{{1}}\right)}+{z}_{{y}}{\left({y}-{y}_{{1}}\right)}\)

\(\displaystyle{z}-{18}={8}{\left({x}-{3}\right)}+{12}{\left({y}-{\left(-{2}\right)}\right)}\)

\(\displaystyle{z}-{18}={8}{x}+{12}{y}\)

\(\displaystyle{z}={8}{x}+{12}{y}+{18}\)

\(\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},{\left({3},-{2},{18}\right)}\)

\(\displaystyle\Rightarrow{z}_{{x}}={4}{\left({x}-{1}\right)}\)

\(\displaystyle={4}{x}-{4}\),

\(\displaystyle{z}_{{y}}={12}{\left({y}+{3}\right)}\)

\(\displaystyle={12}{y}+{36}\)

At the point (3,-2,18):

\(\displaystyle{z}_{{x}}={4}{x}-{4}\)

\(\displaystyle={4}{\left({3}\right)}-{4}\)

\(\displaystyle={12}-{4}\)

\(\displaystyle={8}\)

\(\displaystyle{z}_{{y}}={12}{y}+{36}\)

\(\displaystyle=-{12}{\left(-{2}\right)}+{36}\)

\(\displaystyle=-{24}+{36}\)

\(\displaystyle={12}\)

Equation of the tangent plane is,

\(\displaystyle{z}-{z}_{{1}}={z}_{{x}}{\left({x}-{x}_{{1}}\right)}+{z}_{{y}}{\left({y}-{y}_{{1}}\right)}\)

\(\displaystyle{z}-{18}={8}{\left({x}-{3}\right)}+{12}{\left({y}-{\left(-{2}\right)}\right)}\)

\(\displaystyle{z}-{18}={8}{x}+{12}{y}\)

\(\displaystyle{z}={8}{x}+{12}{y}+{18}\)