Chapter 3.2, Problem 49E

To determine

To find : the two points where the curve crosses the x − axis and show that tangents at these points are parallel.

Answer to Problem 49E

The intersections of curve on x − axis is $\left(-\sqrt{7},0\right)$ and $\left(\sqrt{7},0\right)$ .

The common slope is $m=-2$ .

Explanation of Solution

Given information :

The curve is $x^2+xy+y^2=7$ .

Formula needed :

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

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

Tangent line to curve is $y-y_1=m\left(x-x_1\right)$ where $m$ is slope of curve and $\left(x_1,y_1\right)$ point of intersection of line and curve.

When the curve intersects the x − axis, $y=0$ .

Substitute $y=0$ in the curve $x^2+xy+y^2=7$ .

  $\begin{array}{c}{x}^2+x\cdot 0+0^2=7\\ x^2+0+0=7\\ x^2=7\end{array}$

Take square root on both sides,

  $\begin{array}{c}\sqrt{x^2}=\pm \sqrt{7}\\ x=-\sqrt{7},\sqrt{7}\end{array}$

The intersections of curve on x − axis is $\left(-\sqrt{7},0\right)$ and $\left(\sqrt{7},0\right)$ .

For slope, differentiate $x^2+xy+y^2=7$ with respect to $x$ .

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

Solve further,

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

Plug point $\left(-\sqrt{7},0\right)$ in $\frac{dy}{dx}=-\frac{\left(2x+y\right)}{\left(x+2y\right)}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{\left(-\sqrt{7},0\right)}\\ =-\frac{\left(2\left(-\sqrt{7}\right)+0\right)}{\left(-\sqrt{7}+2\cdot 0\right)}\\ =\frac{2\sqrt{7}}{-\sqrt{7}}\\ =-2\end{array}$

The tangent line to point $\left(-\sqrt{7},0\right)$ is given by $y-0=-2\left(x+\sqrt{7}\right)$ .

Hence, $y=-2x-2\sqrt{7}$ is the tangent line at point $\left(-\sqrt{7},0\right)$ .

Plug point $\left(\sqrt{7},0\right)$ in $\frac{dy}{dx}=-\frac{\left(2x+y\right)}{\left(x+2y\right)}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{\left(\sqrt{7},0\right)}\\ =-\frac{\left(2\left(\sqrt{7}\right)+0\right)}{\left(\sqrt{7}+2\cdot 0\right)}\\ =-\frac{2\sqrt{7}}{\sqrt{7}}\\ =-2\end{array}$

The tangent line to point $\left(\sqrt{7},0\right)$ is given by $y-0=-2\left(x-\sqrt{7}\right)$ .

Hence, $y=-2x+2\sqrt{7}$ is the tangent line at point $\left(\sqrt{7},0\right)$ .

Since, the slope of both tangent line is same. So, the lines are parallel to each other.

Hence, the common slope is $m=-2$ .