Show by implicit differentiation that the tangent line to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_0, y_0)$ has the following equation.
$\frac{x_0x}{a^2} + \frac{y_0y}{b^2} = 1$
Differentiating the left and right-hand sides of the equation of the ellipse with respect to $x$, we have the following.
$\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y' = -\frac{b^2x}{a^2y}$
Therefore, an equation of the tangent line at the point $(x_0, y_0)$ is $y - y_0 = (\text{ })(x - x_0)$.
Multiplying both sides of this equation by $\frac{y_0}{b^2}$ gives the following.
$\frac{y_0y}{b^2} - \frac{y_0^2}{b^2} = \text{ } + \frac{x_0^2}{a^2}$
Since $(x_0, y_0)$ lies on the ellipse, we have the following.
$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$
