Answer
We parametrize the ellipse and find the slope of the tangent line. Then use the equation of the tangent line to obtain $Ax - By = 1$, where $A = \frac{{{x_0}}}{{{a^2}}}$ and $B = \frac{{{y_0}}}{{{b^2}}}$.
Work Step by Step
Write $x=a \sec t$ and $y=b \tan t$. Since ${\sec ^2}t - {\tan ^2}t = 1$, the parametrization
$c\left( t \right) = \left( {x,y} \right) = \left( {a\sec t,b\tan t} \right)$
represent the hyperbola ${\left( {\frac{x}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$.
By Theorem 2 of Section 12.1, the slope of the tangent line of $c\left(t\right)$ is $\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}}$.
We have (from Theorem 2 of Section 3.6)
$x = a\sec t$, ${\ \ }$ $x' = a\sec t\tan t$
$y = b\tan t$, ${\ \ }$ $y' = b{\sec ^2}t$
So,
$\frac{{dy}}{{dx}} = \frac{{y'\left( t \right)}}{{x'\left( t \right)}} = \frac{{b{{\sec }^2}t}}{{a\sec t\tan t}} = \frac{{b\sec t}}{{a\tan t}}$
But $x=a \sec t$ and $y=b \tan t$. So, $\frac{{\sec t}}{{\tan t}} = \frac{{x/a}}{{y/b}} = \frac{{bx}}{{ay}}$. Thus,
$\frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}}$
The equation of the tangent line at a point $P = \left( {{x_0},{y_0}} \right)$ is
$y - {y_0} = \frac{{{b^2}x}}{{{a^2}y}}\left( {x - {x_0}} \right)$
${a^2}{y^2} - {a^2}y{y_0} = {b^2}{x^2} - {b^2}x{x_0}$
${b^2}x{x_0} - {a^2}y{y_0} = {b^2}{x^2} - {a^2}{y^2}$
Divide both sides by ${a^2}{b^2}$ gives
$\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}$
But the right-hand side is equal to 1 by the equation: ${\left( {\frac{x}{a}} \right)^2} - {\left( {\frac{y}{b}} \right)^2} = 1$. Thus,
$\frac{{x{x_0}}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1$
Write $A = \frac{{{x_0}}}{{{a^2}}}$ and $B = \frac{{{y_0}}}{{{b^2}}}$. Hence, $Ax - By = 1$.
