Answer
$$\eqalign{
& \frac{1}{2}x + \frac{2}{3}y + 2\sqrt 3 z = - 2 \cr
& x - 2y + 2\sqrt {14} z = 2 \cr} $$
Work Step by Step
$$\eqalign{ & {z^2} - \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} - 1 = 0;\,\,\,\left( {4,3, - \sqrt 3 } \right){\text{ and }}\left( { - 8,9 - \sqrt {14} } \right) \cr & {\text{Let }}F\left( {x,y,z} \right) = {z^2} - \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} - 1 \cr & {\text{Calculate the partial derivatives }}{F_x}\left( {x,y,z} \right){\text{, }}{F_y}\left( {x,y,z} \right){\text{ }} \cr & {\text{and }}{F_z}\left( {x,y,z} \right){\text{ }} \cr & {F_x}\left( {x,y,z} \right) = - \frac{x}{8} \cr & {F_y}\left( {x,y,z} \right) = - \frac{{2y}}{9} \cr & {F_z}\left( {x,y,z} \right) = 2z \cr & {\text{Evaluate at the point }}\left( {4,3, - \sqrt 3 } \right){\text{ and }}\left( { - 8,9 - \sqrt {14} } \right) \cr & {\text{at }}{F_x}\left( {x,y,z} \right),\,{F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr & {F_x}\left( {4,3, - \sqrt 3 } \right) = - \frac{1}{2} \cr & {F_y}\left( {4,3, - \sqrt 3 } \right) = - \frac{2}{3} \cr & {F_z}\left( {4,3, - \sqrt 3 } \right) = - 2\sqrt 3 \cr & and \cr & {F_x}\left( { - 8,9 - \sqrt {14} } \right) = 1 \cr & {F_y}\left( { - 8,9 - \sqrt {14} } \right) = - 2 \cr & {F_z}\left( { - 8,9 - \sqrt {14} } \right) = - 2\sqrt {14} \cr & \cr & {\text{An equation of the plane tangent to the surface is}} \cr & {F_x}\left( {a,b,c} \right)\left( {x - a} \right) + {F_y}\left( {a,b,c} \right)\left( {y - b} \right) + {F_z}\left( {a,b,c} \right)\left( {z - c} \right) = 0 \cr & {\text{For the point }}\left( {4,3, - \sqrt 3 } \right) \cr & - \frac{1}{2}\left( {x - 4} \right) - \frac{2}{3}\left( {y - 3} \right) + - 2\sqrt {14} \left( {z + \sqrt 3 } \right) = 0 \cr & \frac{1}{2}x + \frac{2}{3}y + 2\sqrt 3 z = - 2 \cr &
{\text{For the point }}\left( { - 8,9 - \sqrt {14} } \right) \cr & \left( {x + 8} \right) - 2\left( {y - 9} \right) - 2\sqrt {14} \left( {z + \sqrt {14} } \right) = 0 \cr & x + 8 - 2y + 18 - 2\sqrt {14} z - 28 = 0 \cr & x - 2y + 2\sqrt {14} z = 2 \cr} $$
