Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 49

Find the points at which the ellipse $x^2 - xy + y^2 = 3$ crosses the $x$-axis and show that the tangent lines at these points are parallel.

The ellipse crosses the $x$-axis at $y = 0$ so...


$
\begin{equation}
\begin{aligned}

x^2 - xy + y^2 =& 3
\\
\\
x^2 - x(0) + (0)^2 =& 3
\\
\\
x^2 =& 3
\\
\\
x =& \pm \sqrt{3}


\end{aligned}
\end{equation}
$


Taking the derivative of the curve implicitly we have,

$x^2 - xy + y^2 = 3$


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (x^2) - \frac{d}{dx} (xy) + \frac{d}{dx} (y^2) =& \frac{d}{dx} (3)
\\
\\
2x - \left[ \frac{d}{dx} (x) \cdot y + x \cdot \frac{d}{dx} (y) \right] + \frac{d}{dy} (y^2) \frac{dy}{dx} =& 0
\\
\\
2x - \left[ y + x \frac{dy}{dx} \right] + 2y \frac{dy}{dx} =& 0
\\
\\
2x - y - x \frac{dy}{dx} + 2y \frac{dy}{dx} =& 0
\\
\\
\frac{dy}{dx} =& \frac{2x - y}{x - 2y}


\end{aligned}
\end{equation}
$


@ $x = \sqrt{3}$


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{2 (\sqrt{3}) - 0}{\sqrt{3} - 2 (0)}
\\
\\
\frac{dy}{dx} =& \frac{2 \cancel{\sqrt{3}} }{\cancel{\sqrt{3}}}
\\
\\
\frac{dy}{dx} =& 2

\end{aligned}
\end{equation}
$


@ $x = - \sqrt{3}$


$
\begin{equation}
\begin{aligned}

\frac{dy}{dx} =& \frac{2 (-\sqrt{3}) - 0}{-\sqrt{3} - 2 (0)}
\\
\\
\frac{dy}{dx} =& \frac{2 (\cancel{-\sqrt{3})} }{\cancel{-\sqrt{3}}}
\\
\\
\frac{dy}{dx} =& 2

\end{aligned}
\end{equation}
$


The slopes are equal. Therefore, the tangent lines are parallel.