Answer
$\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}$ holds on both standard ellipse and hyperbola:
$\begin{array}{*{20}{c}}
{}&{Ellipse\left( {a > b} \right)}&{Hyperbola}\\
{Standard {\ } eq.}&{{{\left( {\frac{x}{a}} \right)}^2} + {{\left( {\frac{y}{b}} \right)}^2} = 1}&{{{\left( {\frac{x}{a}} \right)}^2} - {{\left( {\frac{y}{b}} \right)}^2} = 1}\\
{Derivatives}&{\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}}&{\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}}
\end{array}$
Work Step by Step
By Theorem 1 and Theorem 2 of Section 12.5, the equation of the ellipse and hyperbola in standard position is
Eq. (5) ${\ \ \ }$ ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$
and
Eq. (7) ${\ \ \ }$ ${\left( {\frac{x}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$, ${\ \ \ }$ respectively.
Taking the derivative of $y$ with respect to $x$ gives
Case 1: ellipse
$\frac{2}{a}\left( {\frac{x}{a}} \right) + \frac{2}{b}\left( {\frac{y}{b}} \right)\frac{{dy}}{{dx}} = 0$, ${\ }$ $\frac{{dy}}{{dx}} = \frac{{ - 2x/{a^2}}}{{2y/{b^2}}}$, ${\ }$ $\frac{{dy}}{{dx}} = - \frac{x}{y}\frac{{{b^2}}}{{{a^2}}}$
Consider $a>b>0$.
By Theorem 4, $e = \frac{c}{a} = \frac{{\sqrt {{a^2} - {b^2}} }}{a} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} $.
So, $\frac{{{b^2}}}{{{a^2}}} = 1 - {e^2}$.
Thus, for the ellipse: $\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}$.
Case 2: hyperbola
$\frac{2}{a}\left( {\frac{x}{a}} \right) - \frac{2}{b}\left( {\frac{y}{b}} \right)\frac{{dy}}{{dx}} = 0$, ${\ }$ $\frac{{dy}}{{dx}} = \frac{{2x/{a^2}}}{{2y/{b^2}}}$, ${\ }$ $\frac{{dy}}{{dx}} = \frac{x}{y}\frac{{{b^2}}}{{{a^2}}}$
By Theorem 4, $e = \frac{c}{a} = \frac{{\sqrt {{a^2} + {b^2}} }}{a} = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} $.
So, $\frac{{{b^2}}}{{{a^2}}} = {e^2} - 1$.
Thus, for the hyperbola: $\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}$.
In summary:
$\begin{array}{*{20}{c}}
{}&{Ellipse\left( {a > b} \right)}&{Hyperbola}\\
{Standard {\ } eq.}&{{{\left( {\frac{x}{a}} \right)}^2} + {{\left( {\frac{y}{b}} \right)}^2} = 1}&{{{\left( {\frac{x}{a}} \right)}^2} - {{\left( {\frac{y}{b}} \right)}^2} = 1}\\
{Derivatives}&{\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}}&{\frac{{dy}}{{dx}} = \left( {{e^2} - 1} \right)\frac{x}{y}}
\end{array}$
