Answer
$\frac{x}{4}+y-z=2$
Work Step by Step
$z=\frac{x}{y^{2}}$, $(-4,2-1)$
Consider $f(x,y)= \frac{x}{y^{2}}$
$f_{x}(x,y)=\frac{1}{y^{2}}$
$f_{y}(x,y)=-\frac{2x}{y^{3}}$
At $(-4,2-1)$
$f_{x}(-4,2)=\frac{1}{2^{2}}=\frac{1}{4}$
$f_{y}(-4,2)=-\frac{2\times (-4)}{2^{3}}=1$
The equation of the tangent plane to the given surface at the specified point $(-4,2-1)$ 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
$z+1=\frac{1}{4}(x+4)+1(y-2)$
$z=\frac{x}{4}+1+y-2-1$
Hence, $\frac{x}{4}+y-z=2$
