Chapter 4.2, Problem 50E

(a)

To determine

To find: The pointswhere the tangent is parallel to the x- axis.

Answer to Problem 50E

The points which make the tangent parallel to the x- axis are $x=\pm \sqrt{\frac{7}{3}}$.

Explanation of Solution

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

Calculation:

The points where the tangent is parallel to the x- axis are the points where $\frac{dy}{dx}=0$ .

The derivative of the curve is,

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

Equate $\frac{dy}{dx}=0$ ,

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

The tangent is parallel to the x- axis at $y=-2x$ .

Substitute $y=-2x$ in the curve to find the points.

  $\begin{array}{c}{x}^2+xy+y^2=7\\ x^2+x\left(-2x\right)+{\left(-2x\right)}^2=7\\ x^2-2x^2+4x^2=7\\ 3x^2=7\\ x^2=\frac{7}{3}\\ x=\pm \sqrt{\frac{7}{3}}\end{array}$

Hence, the points are $x=\pm \sqrt{\frac{7}{3}}$ .

(b)

To determine

To find: The points where the tangent is parallel to the y- axis.

Answer to Problem 50E

The points which make the tangent parallel to the y- axis are $y=\pm \sqrt{\frac{7}{3}}$.

Explanation of Solution

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

Calculation:

The points where the tangent is parallel to the y- axis are the points where $\frac{dx}{dy}=0$ .

The derivative of the curve is,

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

Equate $\frac{dx}{dy}=0$ ,

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

The tangent is parallel to the y- axis at $x=-2y$ .

Substitute $x=-2y$ in the curve to find the points.

  $\begin{array}{c}{x}^2+xy+y^2=7\\ {\left(-2y\right)}^2+\left(-2y\right)y+y^2=7\\ 4y^2-2y^2+y^2=7\\ 3y^2=7\\ y^2=\frac{7}{3}\\ y=\pm \sqrt{\frac{7}{3}}\end{array}$

Hence, the points are $y=\pm \sqrt{\frac{7}{3}}$ .