Chapter 3.2, Problem 56E

To determine

To find : the point of intersection of normal line and curve other than $\left(1,1\right)$ .

Answer to Problem 56E

The other point of intersection of normal line and curve is $\left(3,-1\right)$ .

Explanation of Solution

Given information :

The curve is $x^2+2xy-3y^2=0$ and points $\left(1,1\right)$ .

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 $y-b=-\frac{1}{m}\left(x-a\right)$ .

For slope, differentiate $x^2+2xy-3y^2=0$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(x^2+2xy-3y^2\right)=0\\ \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(2xy\right)-\frac{d}{dx}\left(3y^2\right)=0\\ 2x+2y\frac{d}{dx}\left(x\right)+2x\frac{d}{dx}\left(y\right)-6y\frac{dy}{dx}=0\\ 2\left(x+y\right)+2\left(x-3y\right)\frac{dy}{dx}=0\end{array}$

Solving further,

  $\frac{dy}{dx}=-\frac{\left(x+y\right)}{\left(x-3y\right)}$

For slope at $\left(1,1\right)$ , plug $\left(1,1\right)$ in ${\left.\frac{dy}{dx}\right|}_{\left(1,1\right)}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{\left(1,1\right)}\\ =-\frac{\left(1+1\right)}{\left(1-3\cdot 1\right)}\\ =-\frac{2}{-2}\\ =1\end{array}$

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

  $\begin{array}{c}y-1=-x+1\\ y=-x+2\end{array}$

Substitute $y=-x+2$ in the curve $x^2+2xy-3y^2=0$ .

  $\begin{array}{c}{x}^2+2x\left(-x+2\right)-3{\left(-x+2\right)}^2=0\\ x^2+\left(-2x^2+4x\right)-3\left(x^2-4x+4\right)=0\\ x^2-2x^2+4x-3x^2+12x-12=0\\ -4x^2+16x-12=0\end{array}$

Solving further,

  $\begin{array}{c}-4\left(x^2-4x+3\right)=0\\ x^2-4x+3=0\\ \left(x-3\right)\left(x-1\right)=0\\ x=1,3\end{array}$

When $x=1$ , $y=1$ and when $x=3$ , $y=-1$ .

The other point of intersection of normal line and curve is $\left(3,-1\right)$ .