Chapter 4.2, Problem 49E

To determine

To find: The two points where the curve crosses the x- axis. Further, show that the tangents to the curve at these points are parallel, common slope of these tangents.

Answer to Problem 49E

The two points where the curve crosses the x- axis are $x=\sqrt{7}\text{ and }x=-\sqrt{7}$ .

The common slope of these tangents is $\frac{dy}{dx}=-2$ .

Explanation of Solution

Given Information: The equation of the curve is $x^2+xy+y^2=7$

Calculation:

The points where the curve crosses the x- axis can be obtained by finding thex -intercepts. Put $y=0$ . The points are,

  $\begin{array}{c}{x}^2+xy+y^2=7\\ x^2+x\left(0\right)+{\left(0\right)}^2=7\\ x^2=7\\ x=\pm \sqrt{7}\end{array}$

The two points where the curve crosses the x- axis are $x=\sqrt{7}\text{ and }x=-\sqrt{7}$ .

Find the derivative of the function to find the tangents,

  $\begin{array}{c}{x}^2+xy+y^2=7\\ 2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\\ 2x+y+\left(x+2y\right)\frac{dy}{dx}=0\\ \left(x+2y\right)\frac{dy}{dx}=-\left(2x+y\right)\\ \frac{dy}{dx}=-\frac{2x+y}{x+2y}\end{array}$

The tangent at the point $\left(\sqrt{7},0\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2x+y}{x+2y}\\ =-\frac{2\sqrt{7}+0}{\sqrt{7}+0}\\ =-2\end{array}$

The tangent at the point $\left(-\sqrt{7},0\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2x+y}{x+2y}\\ =-\frac{2\left(-\sqrt{7}\right)+0}{\left(-\sqrt{7}\right)+0}\\ =-\frac{2\left(-\sqrt{7}\right)}{\left(-\sqrt{7}\right)}\\ =-2\end{array}$

Clearly, at both the points tangent is equal to -2.

The common slope of these tangents is $\frac{dy}{dx}=-2$ .