91Ó°ÊÓ

Given Information, $f(x,y)=\frac{x}{x^2+y^2−1}$at position $P=\left(x_0,y_0\right)=(1,−3)$

The line of tangent equation is

$f_x\left(x_0,y_0\right)\left(x−x_0\right)+f_y\left(x_0,y_0\right)\left(y−y_0\right)=z−f\left(x_0,y_0\right)$

Equation $1$

$f_x(1,−3)(x−1)+f_y(1,−3)(y+3)=z−f(1,−3)$

Then,

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\frac{x}{x^2+y^2−1}\right|}_{\left(x_0,y_0\right)}$

localid="1650518661208" $f_x\left(x_0,y_0\right)=\frac{\frac{\mathrm{∂}}{\mathrm{∂}x}\left(x\right)\left(x^2+y^2−1\right)−\frac{\mathrm{∂}}{\mathrm{∂}x}\left(x^2+y^2-1)x\right.}{{\left(x^2+y^2−1\right)}^2}$

localid="1650519076415" $f_x\left(x_0,y_0\right)={\left.\frac{1×\left(x^2+y^2−1\right)−2xx}{{\left(x^2+y^2−1\right)}^2}\right|}_{\left(x_0â‹\dots y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.\frac{−x^2+y^2−1}{{\left(x^2+y^2−1\right)}^2}\right|}_{\left(x_0,y_0\right)}$

$f_x(1,−3)={\left.\frac{−x^2+y^2−1}{{\left(x^2+y^2−1\right)}^2}\right|}_{(1,−3)}$

$f_x(1,−3)=\frac{−(1{)}^2+(−3{)}^2−1}{{\left((1{)}^2+(−3{)}^2−1\right)}^2}$

$f_x(1,−3)=\frac{−1+9−1}{1+9−1}$

Equation $2$

$f_x(1,−3)=\frac{7}{9}$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\frac{x}{x^2+y^2−1}\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{∂}}{\mathrm{∂}y}\left(\frac{1}{x^2+y^2−1}\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{∂}}{\mathrm{∂}y}\left({\left(x^2+y^2−1\right)}^{−1}\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{∂}}{\mathrm{∂}y}\left({\left(x^2+y^2−1\right)}^{−1}\right)\frac{\mathrm{∂}}{\mathrm{∂}y}\left(x^2+y^2−1\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\left(−\frac{1}{{\left(x^2+y^2−1\right)}^2}\right)×2y\right|}_{\left(x_0,5_0\right)}$

$f_y(1,−3)=−{\left.\frac{2xy}{{\left(x^2+y^2−1\right)}^2}\right|}_{(1,−3)}$

$f_y(1,−3)=−\frac{2×1×(−3)}{{\left((1{)}^2+(−3{)}^2−1\right)}^2}$

$f_y(1,−3)=−\frac{−6}{1+9−1}$

Equation $3$

$f_y(1,−3)=\frac{6}{9}$

$f\left(x_0,y_0\right)=f(1,−3)=\frac{1}{1^2+(−3{)}^2−1}$

Equation $4$

$f(−3,0)=\frac{1}{9}$

Equation $2,3$and $4$are substituted in equation $1$, we get,

$\frac{7}{9}(x−1)+\frac{6}{9}(y+3)=z−\frac{1}{9}$

$\frac{7}{9}x−\frac{7}{9}+\frac{6}{9}y+\frac{18}{9}=z−\frac{1}{9}$

$\frac{7}{9}x+\frac{6}{9}y−z=−\frac{1}{9}+\frac{7}{9}−\frac{18}{9}$

$\frac{7}{9}x+\frac{6}{9}y−z=−\frac{12}{9}$

$9$is multiplied by two sides, we get,

$7x+6y−9z=−12$

Finally, the result is $7x+6y−9z=−12$.