Find an equation of the tangent line to the graph of the ellipse $3x^{2}+4y^{2}=2x$ at the point $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$.

Solution First we find the slope of the tangent line. We use the result from Example 2, and evaluate $\dfrac{dy}{dx}=\dfrac{1-3x}{4y}$ at $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$. $$\begin{eqnarray*} && \dfrac{dy}{dx}=\dfrac{1-3x}{4y} \underset{\underset{\color{#0066A7}{x=\tfrac{1}{2},y=-\dfrac{1}{4}}}{\color{#0066A7}{{\left\uparrow{\vphantom{\vrule width0pc height12.5pt depth0pt}}\right.}}}}{=} \dfrac{1-3\cdot \dfrac{1}{2}}{4\,{\cdot}\,\!\left(-\dfrac{1}{4}\right) }=\dfrac{1}{2}\\ \end{eqnarray*}$$

The slope of the tangent line to the graph of $3x^{2}+4y^{2}=2x$ at the point $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$ is $\dfrac{1}{2}$. An equation of the tangent line is $$\begin{eqnarray*} y+\dfrac{1}{4} &=&\dfrac{1}{2}\!\left( x-\dfrac{1}{2}\right) \\ y &=&\dfrac{1}{2}x-\dfrac{1}{2} \end{eqnarray*}$$