Using Implicit Differentiation to Find an Equation of a Tangent Line

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 \[ \begin{eqnarray*} y+\dfrac{1}{4} &=&\dfrac{1}{2}\!\left( x-\dfrac{1}{2}\right) \\ y &=&\dfrac{1}{2}x-\dfrac{1}{2} \end{eqnarray*} \]