The problem asks for the standard form of a hyperbola given the following details:

• Center: $(0,0)$
• Transverse axis: Along the $x$-axis (horizontal hyperbola)
• Asymptotes: $y = 4x$ and $y = -4x$

Step-by-step solution:

1. General form of a horizontal hyperbola centered at $(0, 0)$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

   • The transverse axis is along the $x$-axis, so the $x^2$-term comes first.
   • $a$ and $b$ are positive constants representing distances from the center to vertices and co-vertices, respectively.

2. Relationship between asymptotes and hyperbola parameters: The asymptotes of a horizontal hyperbola are given by the equations: $y = \pm \frac{b}{a}x$ For this problem, the slopes of the asymptotes are $\pm 4$, meaning: $\frac{b}{a} = 4$ This implies: $b = 4a$

3. Substitute $b = 4a$ into the standard form: Now we rewrite the equation using the relationship we found.

   • Since $b^2 = (4a)^2 = 16a^2$, the equation becomes: $\frac{x^2}{a^2} - \frac{y^2}{16a^2} = 1$

4. Eliminate the common factor $a^2$: Divide both terms by $a^2$: $\frac{x^2}{a^2} - \frac{y^2}{16a^2} = 1 \quad \Longrightarrow \quad \frac{x^2}{1} - \frac{y^2}{16} = 1$

5. Final standard form: $\boxed{x^2 - \frac{y^2}{16} = 1}$

This is the standard equation of the hyperbola with the given center, transverse axis, and asymptotes.

Would you like me to break down any part of the solution further or clarify a specific step?

Here are five follow-up questions related to this topic:

1. How do you determine whether a hyperbola is horizontal or vertical?
2. What changes in the equation if the hyperbola's center is at $(h, k)$ instead of $(0,0)$?
3. How do you find the foci of the hyperbola from its equation?
4. What role do the transverse and conjugate axes play in defining a hyperbola?
5. Can two different hyperbolas have the same asymptotes?

Tip: The slopes of the asymptotes are always determined by the ratio $\pm \frac{b}{a}$, so identifying them early simplifies the process.