In Exercises $29–34,$find the equation of the line tangent to the

surface at the given point P and in the direction of the given

unit vector u. Note that these are the same functions, points,

and vectors as in Exercises$21–26.$

$f(x,y)=\frac{x}{y^2}\text{at}P=(-2,1),\mathbf{u}=-\frac{5}{13}\mathbf{i}+\frac{12}{13}\mathrm{j}$

$f(x,y)=\frac{x}{y^2}$

At point$P=\left(x_0,y_0\right)=(-2,1)\text{and}u=-\frac{5}{13}i+\frac{12}{13}j$

The equation of line tangent is

$\begin{array}{r}{f}_x\left(x_0,y_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0\right)\left(y-y_0\right)=z-f\left(x_0,y_0\right)\end{array}$

$\begin{array}{r}{f}_x(-2,1)(x+2)+f_y(-2,1)(y-1)=z-f(-2,1)\end{array}$ $\left(\left(1\right)\right)$

Assume

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0{5}_0\right)}={\left.\frac{d}{dx}\frac{x}{y^2}\right|}_{\left(x_0,v_0\right)}$

$f_x(-2,1)={\left.\frac{1}{y^2}\right|}_{(-2,1)}$

$f_x(-2,1)=1$ $\left(2\right)$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\frac{x}{y^2}\right|}_{\left(x_0,y_0\right)}$

$f_y(-2,1)=4$

fy(−2,1)=4

$f\left(x_0,y_0\right)=f(-2,1)=\frac{-2}{1}$

$f(-2,1)=-2$

Substituting

$1(x+2)+4(y-1)=z+2$

$x+2+4y-4=z+2$

$x+4y-z=2-2+4$

$x+4y-z=4$

Consider the equation of normal line

here

$x=x_0+\alpha t,y=y_0+\beta t,z=z_0+\gamma t$

where

$z_0=f\left(x_0,y_0\right)\text{and}\gamma =D_{w}f\left(x_0,y_0\right)$

Consider directional derivative

$D_{w}f\left(x_0,y_0\right)={\mathrm{Lim}}_{h\to 0}\frac{f\left(x_0+\alpha h,y_0+\beta h\right)-f\left(x_0,y_0\right)}{h}$

$D_{w}f(-2,1)={\mathrm{Lim}}_{h\to 0}\frac{f\left(-2-\frac{5}{13}h,1+\frac{12}{13}h\right)-f(-2,1)}{h}$

$f\left(-2-\frac{5}{13}h,1+\frac{12}{13}h\right)=\frac{-2-\frac{5}{13}h}{{\left(1+\frac{12}{13}h\right)}^2}$

$=\frac{-2-\frac{5}{13}h}{\left(1+\frac{24}{13}h+\frac{144}{169}{h}^2\right)}$

$\frac{\frac{-26-5h}{13}}{\left.\frac{24}{13}h+\frac{144}{169}{h}^2\right)}$

$=\frac{-26-5h}{13\times \left(\frac{169+312h+144h^2}{169}\right)}$

$=\frac{-338-65h}{169+312h+144h^2}$ $\left(6\right)$

and

$f\left(x_0,y_0\right)=f(-2,1)=\frac{-2}{1^2}=-2$ $\left(7\right)$

$={\mathrm{Lim}}_{h\to 0}\frac{-338-65h+338+624h+288h^2}{h\left(169+312h+144h^2\right)}$

$={\mathrm{Lim}}_{h\to 0}\frac{559h+288h^2}{h\left(169+312h+144h^2\right)}$

$={\mathrm{Lim}}_{h\to 0}\frac{h(559+288h)}{h\left(169+312h+144h^2\right)}$

role="math" localid="1650363916684" $D_{N}f(-2,1)=\frac{43}{13}$

Therefore,

role="math" localid="1650363905575" $x=-2-\frac{5}{13}t,y=1+\frac{12}{13}t,z=-2+\frac{43}{13}t$