Chapter 3.2, Problem 57E

To determine

To find : the normal line which is parallel to $2x+y=0$ .

Answer to Problem 57E

The normal equation to curve at $\left(-1,-1\right)$ is given by $y=-2x-3$ .

The normal equation to curve at $\left(3,-3\right)$ is given by $y=-2x+3$ .

Explanation of Solution

Given information :

The curve is $xy+2x-y=0$ and line parallel to normal line is $2x+y=0$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

Power rule of derivative: $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

The normal equation to curve at $\left(a,b\right)$ is given by $y-b=-\frac{1}{m}\left(x-a\right)$ .

Compare $y-b=-\frac{1}{m}\left(x-a\right)$ with $2x+y=0$ .

  $\begin{array}{c}-\frac{1}{m}=-2\\ m=\frac{1}{2}\end{array}$

For slope, differentiate $xy+2x-y=0$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(xy+2x-y=0\right)=0\\ \frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(y\right)=0\\ x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(x\right)+2-\frac{dy}{dx}=0\\ x\frac{dy}{dx}+y+2-\frac{dy}{dx}=0\end{array}$

Solving further,

  $\begin{array}{c}\left(x-1\right)\frac{dy}{dx}+\left(y+2\right)=0\\ \left(x-1\right)\frac{dy}{dx}=-\left(y+2\right)\\ \frac{dy}{dx}=-\frac{\left(y+2\right)}{\left(x-1\right)}\end{array}$

Take both slopes equally.

  $\begin{array}{c}-\frac{\left(y+2\right)}{\left(x-1\right)}=\frac{1}{2}\\ -2\left(y+2\right)=\left(x-1\right)\\ -2y-4=x-1\\ y=\frac{1}{-2}\left(x+3\right)\end{array}$

Plug above equation in the curve.

  $\begin{array}{c}x\left(\frac{1}{-2}\left(x+3\right)\right)+2x-\left(\frac{1}{-2}\left(x+3\right)\right)=0\\ -\frac{x^2}{2}-\frac{3x}{2}+2x+\frac{x}{2}+\frac{3}{2}=0\\ -\frac{x^2}{2}+x+\frac{3}{2}=0\\ -x^2+2x+3=0\\ x=-1,3\end{array}$

The coordinates are $\left(-1,-1\right)$ and $\left(3,-3\right)$ .

The normal equation to curve at $\left(-1,-1\right)$ is given by $y+1=-2\left(x+1\right)$ .

The normal equation to curve at $\left(3,-3\right)$ is given by $y+3=-2\left(x-3\right)$ .