Implicit Differentiation - Exercise 2


Exercise 2. Prove that an equation of the tangent line to the graph of the hyperbola

\begin{displaymath}\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\end{displaymath}

at the point P(x0,y0) is

\begin{displaymath}\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1\end{displaymath}

Answer. Instead of finding y explicitly as a function of x, we will use implicit differentiation to find the slope of the tangent line. We have

\begin{displaymath}2 \frac{x}{a^2} - \frac{2 y y'}{b^2} = 0\end{displaymath}

or equivalently

\begin{displaymath}y' = \frac{x/a^2}{y/b^2} \;\cdot\end{displaymath}

So the equation of the tangent line at the point P(x0,y0) is

\begin{displaymath}y - y_0 = \frac{x_0/a^2}{y_0/b^2} (x-x_0)\end{displaymath}

or equivalently

\begin{displaymath}\frac{y_0}{b^2} (y - y_0) = \frac{x_0}{a^2} (x-x_0)\;.\end{displaymath}

Knowing that

\begin{displaymath}\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1\end{displaymath}

the equation of the tangent line becomes

\begin{displaymath}\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1\;.\end{displaymath}


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