Chapter 4.2, Problem 56E

To determine

To find: The other point at which a line intersects the curve $x^2+2xy-3y^2=0$

Answer to Problem 56E

The other point is $\left(3,-1\right)$ .

Explanation of Solution

Given Information: One point is $\left(1,1\right)$ .

Calculation:

The derivative $\frac{dy}{dx}$ of the curve is,

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

The slope at the point $\left(1,1\right)$ is,

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

This is the slope of the tangent. The slope of the normal is,

  $\begin{array}{c}{m}_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}\\ =-\frac{1}{1}\\ =1\end{array}$

• Here, $m_{\text{normal}}$ is the slope of the normal line and $m_{\text{tangent}}$ is the slope of the tangent line.

The equation of the line of normal is,

  $\begin{array}{c}y-1=m_{\text{normal}}\left(x-1\right)\\ y-1=-1\left(x-1\right)\\ y-1=-x+1\\ y=-x+2\end{array}$

Substitute $y=-x+2$ in $x^2+2xy-3y^2=0$ ,

  $\begin{array}{c}{x}^2+2xy-3y^2=0\\ x^2+2x\left(-x+2\right)-3{\left(-x+2\right)}^2=0\\ x^2-2x^2+4x-3\left(x^2+4-4x\right)=0\\ x^2-2x^2+4x-3x^2-12+12x=0\\ -4x^2+16x-12=0\\ x^2-4x+3=0\\ x^2-x-3x+3=0\\ x\left(x-1\right)-3\left(x-1\right)=0\\ \left(x-1\right)\left(x-3\right)=0\\ x=1,3\end{array}$

The corresponding y- coordinates are,

  $\begin{array}{l}\text{When }x=1,y=-1+2=1\\ \text{When }x=3,y=-3+2=-1\end{array}$

Hence, the required coordinates are $\left(1,1\right)\text{ and }\left(3,-1\right)$ .

Since, $\left(1,1\right)$ point is already given, the other point is $\left(3,-1\right)$ .

Conclusion:

The other point is $\left(3,-1\right)$ .