4. An engineer is designing a diagram for a cooling tower he plans to construct in the future. Cooling towers often have sides that curve inward, resembling the shape of a hyperbola. His daughter asks him how high the center of the hyperbolic sides of the tower is, and he responds that the center can be situated at (0,15) on the Cartesian plane. The engineer then mentions that the asymptotes of the hyperbola are 𝑦=±24/7 𝑥+15, and the area of its auxiliary rectangle is 168 square units. The daughter then calculates the foci of the hyperbola. At what points on the Cartesian plane are these foci located?

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 \((0, 2.5)\).

---

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!

An engineer is designing a diagram for a cooling tower he plans to construct in the future. Cooling towers often have sides that curve inward, resembling the shape of a hyperbola. His daughter asks him how high the center of the hyperbolic sides of the tower is, and he responds that the center can be situated at (0,15) on the Cartesian plane. The engineer then mentions that the asymptotes of the hyperbola are 𝑦=±24/7 𝑥+15, and the area of its auxiliary rectangle is 168 square units. The daughter then calculates the foci of the hyperbola. At what points on the Cartesian plane are these foci located?

Explore how to calculate the foci of a hyperbola with center (0, 15), given asymptotes y = ±24/7x + 15 and an auxiliary rectangle area of 168 square units. Learn key concepts like the equation of a hyperbola, slope of asymptotes, and calculating the distance to the foci. Suitable for Grades 11-12 students.

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