Answer
$z=4x-y-6$
or
$4x-y-z=6$
Work Step by Step
$z=2x^{2}+y^{2}-5y$, $(1,2,-4)$
Consider $f(x,y)= 2x^{2}+y^{2}-5y$
$f_{x}(x,y)=4x$
$f_{y}(x,y)=2y-5$
At $(1,2,-4)$
$f_{x}(-4,2)=4\times1=4$
$f_{y}(-4,2)=2\times 2 -5=-1$
The equation of the tangent plane to the given surface at the specified point $(1,2,-4)$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
$z+4=4(x-1)+(-1)(y-2)$
$z=4x-4-y+2-4$
Hence, $z=4x-y-6$ or $4x-y-z=6$
