91Ó°ÊÓ

Given line target is,

Point of,$P=\left(x_0,y_0\right)=(4,9)$and $\mathrm{u}=(\mathrm{Î\pm },\mathrm{β})=\left(−\frac{\sqrt{17}}{17},−\frac{4\sqrt{17}}{17}\right)$

The equation of line of tangent is

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

$f_x(4,9)(x−4)+f_y(4,9)(y−9)=z−f(4,9)$

Regarded,

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,0_0\right)}={\left.\frac{d}{dx}\sqrt{\frac{y}{x}}\right|}_{\left(x_0,y_0\right)}$

$f_x(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{\left(4\right)}={\left.\frac{\frac{−y}{x^2}}{2\sqrt{\frac{y}{x}}}\right|}_{\left(40\right)}$

$f_x(4,9)=\frac{−9}{4^2}$

$f_x(4,9)=−\frac{9}{\frac{9}{4}}$

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

$f_y(4,9)={\left.\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right|}_{\left(4\right)}=\frac{\frac{1}{x}}{{\left.2\sqrt{\frac{y}{x}}\right|}_{\left(49\right)}}$

$f_x(4,9)=\frac{1}{2\sqrt{\frac{9}{4}}}$

$f_x(4,9)=\frac{1}{12}$ $\left(3\right)$

$f\left(x_0,y_0\right)=f(4,9)=\sqrt{\frac{9}{4}}$

$f(−2,1)=\frac{3}{2}$ $\left(4\right)$

Substituting

$\frac{−9}{48}(x−4)+\frac{1}{12}(y−9)=z−\frac{3}{2}$

$\frac{−9}{48}x+\frac{36}{48}+\frac{1}{12}y−\frac{8}{12}=z−\frac{3}{2}$

$\frac{−3}{16}x+\frac{3}{4}+\frac{1}{12}y−\frac{2}{3}=z−\frac{3}{2}$

$\frac{−3}{16}x+\frac{1}{12}y−z=−\frac{3}{2}−\frac{3}{4}+\frac{2}{3}$

Multiply $48$on both sides

$−9x+4y−48z=48\left(\frac{−19}{12}\right)$

$−9x+4y−48z=−76$

Consider equation of normal line

Now,

$x=x_0+\mathrm{Î\pm }t,y=y_0+\mathrm{β}t,z=z_0+\mathrm{γ}t$

where

$z_0=f\left(x_0,y_0\right)\text{and}\mathrm{γ}=\mathrm{∇}f\left(P\right)â‹\dots u$

Directional derivative of function at point$P$with directional derivative given by

$\mathrm{∇}f\left(P\right)â‹\dots u=\mathrm{∇}f(4,9)â‹\dots u=\left({\left.\frac{df}{dx}\right|}_{(4,9)}i+{\left.\frac{df}{dy}\right|}_{\left(4.9\right)}j\right)â‹\dots \left(−\frac{\sqrt{17}}{17}i−\frac{4\sqrt{17}}{17}j\right)$

$=\left[{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{\text{(4.9)}}i+{\left(\frac{1}{2\sqrt{\frac{y}{x}}}\frac{d}{dx}\left(\frac{y}{x}\right)\right)}_{(4,9)}j\right]â‹\dots \left(−\frac{\sqrt{17}}{17}i−\frac{4\sqrt{17}}{17}j\right)$

role="math" $=\left[{\left(\frac{\frac{−y}{x^2}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}i+{\left(\frac{\frac{1}{x}}{2\sqrt{\frac{y}{x}}}\right)}_{(4,9)}jâ‹\dots \left(−\frac{\sqrt{17}}{17}i−\frac{4\sqrt{17}}{17}j\right)\right.$

role="math" localid="1650366447035">=−942294i+14294j⋅−1717i−41717j

$=\left[\left(\frac{−\frac{9}{16}}{2â‹\dots \frac{3}{2}}\right)i+\left(\frac{\frac{1}{4}}{2â‹\dots \frac{3}{2}}\right)jâ‹\dots \left(−\frac{\sqrt{17}}{17}i−\frac{4\sqrt{17}}{17}j\right)\right.$

$=\left(−\frac{9}{48}i+\frac{1}{12}j\right)â‹\dots \left(−\frac{\sqrt{17}}{17}i−\frac{4\sqrt{17}}{17}j\right)$

$=\left(\frac{9×\sqrt{17}}{48×17}−\frac{4\sqrt{17}}{12×17}\right)$

$=\frac{1}{3×17}\left(\frac{9×\sqrt{17}}{16}−\frac{\sqrt{17}}{1}\right)$

$=\frac{1}{3×17}\left(\frac{9×\sqrt{17}−16\sqrt{17}}{16}\right)$

$\mathrm{∇}f\left(P\right)â‹\dots u=−\frac{7}{816}\sqrt{17}$

Therefore

$x=4−\frac{\sqrt{17}}{17}t,y=9−\frac{4\sqrt{17}}{17},t,z=\frac{3}{2}−\frac{7\sqrt{17}}{816}t$