The given points (3, -2) and (3, 6) are the endpoints of the conjugate axis of a hyperbola. The conjugate axis is perpendicular to the transverse axis. We are also given that the length of the transverse axis is 16 units.

Step 1: Determine the center of the hyperbola

The center of the hyperbola lies at the midpoint of the conjugate axis.

The midpoint formula is: $\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

For the given points (3, -2) and (3, 6): $\text{Midpoint} = \left( \frac{3 + 3}{2}, \frac{-2 + 6}{2} \right) = (3, 2)$

Thus, the center of the hyperbola is at (3, 2).

Step 2: Use the transverse axis information

The length of the transverse axis is given as 16 units, meaning $2a = 16$. Hence, the value of $a$ is: $a = \frac{16}{2} = 8$

The transverse axis runs along the x-axis (horizontal) because the conjugate axis is along the y-axis (vertical).

Step 3: Write the equation of the hyperbola

Since the transverse axis is horizontal, the equation of the hyperbola takes the form: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ where $(h, k)$ is the center of the hyperbola, and $a$ is half the length of the transverse axis.

Thus, $h = 3$, $k = 2$, and $a = 8$. Therefore, $a^2 = 64$.

Step 4: Determine $b$

The distance between the endpoints of the conjugate axis is 8 units, as the difference in their y-coordinates is $6 - (-2) = 8$. Hence, the length of the conjugate axis is 8 units, so $2b = 8$, and: $b = \frac{8}{2} = 4$

Thus, $b^2 = 16$.

Final Equation:

Substituting the values into the standard equation of the hyperbola: $\frac{(x - 3)^2}{64} - \frac{(y - 2)^2}{16} = 1$

This is the equation of the hyperbola.

Would you like further details or have any questions?

Here are five related questions:

1. How do we identify the transverse and conjugate axes in a hyperbola?
2. What are the differences between hyperbolas and ellipses in terms of axes?
3. How can you derive the equation of a hyperbola if only given the foci?
4. What role does the distance between the vertices play in hyperbola equations?
5. How can the asymptotes of a hyperbola be determined from its equation?

Tip: When working with hyperbolas, the transverse axis is always along the direction of the hyperbola's opening, and the conjugate axis is perpendicular to it.