Answer
$z=8x-8y+11$
Work Step by Step
Given: $z=(x+2)^{2}-2(y-1)^{2}-5$, $(2,3,3)$
Consider $f(x,y)= (x+2)^{2}-2(y-1)^{2}-5$
$f_{x}(x,y)=2(x+2)$
$f_{y}(x,y)=-4(y-1)$
At $(2,3,3)$
$f_{x}(2,3)=8$
$f_{y}(2,3)=-8$
The equation of the tangent plane to the given surface at the specified point $(2,3,3)$ is given by
$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
On substituting the values, we get
$z-3=8(x-2)-8(y-3)$
$z=8x-16-8y+24+3$
Hence, $z=8x-8y+11$
