To solve this problem, we need to determine the equation of the tangent line to the hyperbola:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

at the point $(x_0, y_0)$. Here's the process:

Step 1: Implicit Differentiation

We start by differentiating the given equation implicitly with respect to $x$:

$\frac{d}{dx} \left( \frac{x^2}{a^2} \right) - \frac{d}{dx} \left( \frac{y^2}{b^2} \right) = \frac{d}{dx}(1)$

This gives:

$\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$

Simplify to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{\frac{x}{a^2}}{\frac{y}{b^2}} = \frac{b^2 x}{a^2 y}$

So, the slope of the tangent line ($m$) is:

$m = \frac{dy}{dx} = \frac{b^2 x_0}{a^2 y_0}$

Step 2: Equation of the Tangent Line

The equation of a line is given by the point-slope form:

$y - y_0 = m (x - x_0)$

Substitute $m = \frac{b^2 x_0}{a^2 y_0}$:

$y - y_0 = \frac{b^2 x_0}{a^2 y_0} (x - x_0)$

Simplify:

$y - y_0 = \frac{b^2 x_0 x}{a^2 y_0} - \frac{b^2 x_0^2}{a^2 y_0}$

Reorganize to get:

$\frac{b^2 x_0 x}{a^2 y_0} - y + y_0 = \frac{b^2 x_0^2}{a^2 y_0}$

This is the equation of the tangent line.

Would you like further clarification or assistance? Here are some related questions you might want to explore:

1. How do you compute the slope for tangents to other conic sections (e.g., ellipses)?
2. What happens to the tangent equation when the point $(x_0, y_0)$ is a vertex of the hyperbola?
3. Can you derive the tangent equation for a hyperbola in rotated coordinates?
4. How does the equation change for a horizontal or vertical hyperbola?
5. What are the key differences between tangents to hyperbolas and ellipses?

Tip: Always simplify the equation of the tangent line to make it easier to interpret and use in related problems.