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The given equation represents a hyperbola in the form:

\[
\frac{(x - 4)^2}{64} - \frac{(y - 8)^2}{81} = 1
\]

This is the equation of a hyperbola oriented horizontally, with the center at \((h, k) = (4, 8)\). For hyperbolas, the asymptotes are given by:

\[
y - k = \pm \frac{b}{a}(x - h)
\]

Where:
- \( (h, k) \) is the center of the hyperbola.
- \( a^2 = 64 \) so \( a = \sqrt{64} = 8 \).
- \( b^2 = 81 \) so \( b = \sqrt{81} = 9 \).

Thus, the asymptote equations are:

\[
y - 8 = \pm \frac{9}{8}(x - 4)
\]

The positive slope equation corresponds to the positive sign in the equation:

\[
y - 8 = \frac{9}{8}(x - 4)
\]

To put this in slope-intercept form, we simplify:

\[
y = \frac{9}{8}(x - 4) + 8
\]

Distribute the \(\frac{9}{8}\):

\[
y = \frac{9}{8}x - \frac{9}{2} + 8
\]

Simplifying:

\[
y = \frac{9}{8}x + \frac{7}{2}
\]

So, the equation of the asymptote with a positive slope is:

\[
y = \frac{9}{8}x + \frac{7}{2}
\]

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. How do you find the center of a hyperbola given its equation?
2. What are the general forms of equations for horizontal and vertical hyperbolas?
3. How do you calculate the slope of asymptotes for hyperbolas?
4. What distinguishes a hyperbola from an ellipse in terms of their equations?
5. How do you derive the equation of the asymptote if the hyperbola is vertical instead of horizontal?

**Tip:** For any hyperbola, always check the terms involving \(x\) and \(y\) to determine if it's horizontal or vertical, which determines the direction of the asymptotes!

Find the equation of the asymptote with a positive slope for the following hyperbola: (x - 4)^2 / 64 - (y - 8)^2 / 81 = 1.

Learn how to find the equation of the asymptote with a positive slope for the hyperbola (x - 4)^2 / 64 - (y - 8)^2 / 81 = 1. This problem involves key concepts of conic sections and the asymptotes of hyperbolas. Suitable for students in Grades 10-12.

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