Endpoints of Conjugate axis (3,-2), (3,6), length of the transverse axis is 16 units

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.

Learn how to find the equation of a hyperbola with the endpoints of its conjugate axis at (3,-2) and (3,6) and a transverse axis length of 16 units. This problem explores key concepts in analytic geometry and hyperbolas, using the midpoint formula and the standard hyperbola equation. Ideal for students in Grades 10-12.

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