Chapter 3.5, Problem 49E

To determine

To find: The points when the ellipse crosses the x-axis and to show that the tangent line at these points are parallel.

Answer to Problem 49E

The ellipse crosses the x-axis at $\left(-\sqrt{3},0\right)\text{ and }\left(\sqrt{3},0\right)$ and its tangent lines are parallel.

Explanation of Solution

Given:

The equation of the ellipse $x^2-xy+y^2=3$.

Derivative rules:

Product rule:$\frac{d}{dx}\left(fg\right)=f\frac{d}{dx}\left(g\right)+g\frac{d}{dx}\left(f\right)$

Chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Calculation:

Obtain the points if the ellipse crosses the x-axis.

The ellipse crosses the x-axis. That is, $y=0$.

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

$\begin{array}{l}{x}^2-0+0=3\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

Therefore, the ellipse crosses the x-axis at $\left(-\sqrt{3},0\right)\text{ and }\left(\sqrt{3},0\right)$.

Obtain the slope of the tangent at the points $\left(-\sqrt{3},0\right)\text{ and }\left(\sqrt{3},0\right)$.

Differentiate $x^2-xy+y^2=3$ implicitly with respect to x.

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

Apply the product rule and the chain rule.

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

Separate $\frac{dy}{dx}$ on one side of the equation,

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

Thus, the derivative of the equation is $\frac{dy}{dx}=\frac{y-2x}{2y-x}$.

That is, the slope of the tangent is $\frac{dy}{dx}=\frac{y-2x}{2y-x}$.

The slope of the tangent at $\left(-\sqrt{3},0\right)$ is computed as follows,

$\begin{array}{c}\frac{dy}{dx}=\frac{0-2\left(-\sqrt{3}\right)}{0+\sqrt{3}}\\ =\frac{2\sqrt{3}}{\sqrt{3}}\\ =2\end{array}$

Thus, the slope of the tangent at $\left(-\sqrt{3},0\right)$ is 2.

The slope of the tangent at $\left(\sqrt{3},0\right)$ is computed as follows,

$\begin{array}{c}\frac{dy}{dx}=\frac{0-2\left(\sqrt{3}\right)}{0-\sqrt{3}}\\ =\frac{-2\left(\sqrt{3}\right)}{-\sqrt{3}}\\ =2\end{array}$

Thus, the slope of the tangent at $\left(\sqrt{3},0\right)$ is 2.

Note that, the slope of the tangent lines at the points $\left(-\sqrt{3},0\right)\text{ and }\left(\sqrt{3},0\right)$ are same.

Hence, the tangent lines at $\left(-\sqrt{3},0\right)\text{ and }\left(\sqrt{3},0\right)$ are parallel to the equation is proved.