We are given a hyperbola with its center at $(0, 15)$, and asymptotes described by the equations:

$y = \pm \frac{24}{7} x + 15$

The equation of a hyperbola with vertical transverse axis and center at $(h, k)$ can be written as:

$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

In our case, the center is $(0, 15)$, so the equation of the hyperbola becomes:

$\frac{x^2}{a^2} - \frac{(y - 15)^2}{b^2} = 1$

Step 1: Find the slopes of the asymptotes

The equations of the asymptotes are:

$y = \frac{24}{7}x + 15 \quad \text{and} \quad y = -\frac{24}{7}x + 15$

From this, we can extract the slopes of the asymptotes, which are $\pm \frac{24}{7}$.

The slopes of the asymptotes for a hyperbola are given by $\pm \frac{b}{a}$, where $b$ is the distance from the center to the co-vertices, and $a$ is the distance from the center to the vertices.

Thus, we can equate $\frac{b}{a} = \frac{24}{7}$.

$b = \frac{24}{7}a$

Step 2: Use the area of the auxiliary rectangle

The area of the auxiliary rectangle is given as 168 square units. The area of this rectangle is also $2a \times 2b = 4ab$. Therefore:

$4ab = 168$

Simplifying:

$ab = 42$

Substitute $b = \frac{24}{7}a$ into this equation:

$a \times \frac{24}{7}a = 42$

$\frac{24}{7}a^2 = 42$

Multiply both sides by 7:

$24a^2 = 294$

Solve for $a^2$:

$a^2 = \frac{294}{24} = 12.25$

Thus, $a = \sqrt{12.25} = 3.5$.

Step 3: Find $b$

Now that we know $a = 3.5$, we can find $b$ using the relation $b = \frac{24}{7}a$:

$b = \frac{24}{7} \times 3.5 = 12$

Step 4: Calculate the distance to the foci

The distance from the center to each focus is given by $c$, where $c^2 = a^2 + b^2$. Using the values of $a^2 = 12.25$ and $b^2 = 144$:

$c^2 = a^2 + b^2 = 12.25 + 144 = 156.25$

Thus:

$c = \sqrt{156.25} = 12.5$

Step 5: Find the coordinates of the foci

Since the hyperbola has a vertical transverse axis, the foci are located at $(0, 15 \pm c)$. Therefore, the coordinates of the foci are:

$(0, 15 + 12.5) = (0, 27.5) \quad \text{and} \quad (0, 15 - 12.5) = (0, 2.5)$

Thus, the foci of the hyperbola are located at $(0, 27.5)$ and $(0, 2.5)$.

Summary:

The coordinates of the foci are $(0, 27.5)$ and $(0, 2.5)$.

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Do you need more details on any step or have any questions?

Here are 5 related questions for further exploration:

1. How would the equation of the hyperbola change if the transverse axis were horizontal?
2. What is the significance of the foci in a hyperbola?
3. How do the asymptotes help in sketching the graph of a hyperbola?
4. Can the area of the auxiliary rectangle determine other features of the hyperbola?
5. How do you derive the equation of a hyperbola from its geometric definition?

Tip: For any conic section, asymptotes provide useful boundary behavior that helps in graphing the curve efficiently!