An engineer is working on making a diagram for the cooling tower he was going to con
struct in the future. Cooling towers can be seen as towers with inward-curving sides, or 
simply, hyperbola-looking sides. His daughter asks him how high the center of the hyper
blic sides of the tower was                 , and he answers that the center can be situated at (0,15) in the Cartesian plane. The engineer then says that the asymptotes of the hy
perbola are the following: y=+-24/7x+15, and the area of its auxiliary rectangle is 168 square units. The 
daughter then finds the foci of the hyperbola. At what points on the Cartesian plane are 
these foci found on?

We are given the following details about the hyperbola:

- The center of the hyperbola is at \( (0, 15) \).
- The equations of the asymptotes are \( y = \pm \frac{24}{7}x + 15 \).
- The area of the auxiliary rectangle is 168 square units.

### Step 1: General form of the hyperbola
Since the hyperbola has asymptotes in the form \( y = \pm \frac{m}{n}x + k \), we recognize it as a horizontal hyperbola with the general equation:
\[
\frac{x^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]
where \( (0, k) \) is the center of the hyperbola.

Here, \( k = 15 \), which gives us the center at \( (0, 15) \). We also know the slopes of the asymptotes are \( \pm \frac{24}{7} \), and for a horizontal hyperbola, the slope of the asymptotes is given by \( \pm \frac{b}{a} \). Hence, we can set:
\[
\frac{b}{a} = \frac{24}{7}
\]
This implies:
\[
b = \frac{24}{7}a
\]

### Step 2: Area of the auxiliary rectangle
The area of the auxiliary rectangle is given by \( 2a \times 2b = 4ab \). We are told that this area is 168 square units, so we can set up the equation:
\[
4ab = 168
\]
Simplifying this:
\[
ab = 42
\]
Substitute \( b = \frac{24}{7}a \) into this equation:
\[
a \left( \frac{24}{7}a \right) = 42
\]
\[
\frac{24}{7}a^2 = 42
\]
Multiply both sides by 7 to eliminate the fraction:
\[
24a^2 = 294
\]
Solve for \( a^2 \):
\[
a^2 = \frac{294}{24} = 12.25
\]
Thus, \( a = \sqrt{12.25} = 3.5 \).

### Step 3: Find \( b^2 \)
Now, substitute \( a = 3.5 \) into the equation \( b = \frac{24}{7}a \):
\[
b = \frac{24}{7} \times 3.5 = 12
\]
Thus, \( b^2 = 12^2 = 144 \).

### Step 4: Calculate the distance to the foci
For a hyperbola, the distance from the center to the foci is given by \( c \), where \( c^2 = a^2 + b^2 \). Using the values of \( a^2 = 12.25 \) and \( b^2 = 144 \):
\[
c^2 = 12.25 + 144 = 156.25
\]
Thus, \( c = \sqrt{156.25} = 12.5 \).

### Step 5: Determine the coordinates of the foci
Since the hyperbola is horizontal and centered at \( (0, 15) \), the foci are located at \( (c, 15) \) and \( (-c, 15) \). Therefore, the coordinates of the foci are:
\[
(12.5, 15) \quad \text{and} \quad (-12.5, 15)
\]

### Final Answer:
The foci of the hyperbola are located at \( (12.5, 15) \) and \( (-12.5, 15) \) on the Cartesian plane.

Would you like more details on any part of the solution or have further questions?

---

Here are five related questions:
1. How do the slopes of the asymptotes affect the shape of a hyperbola?
2. How is the equation of the hyperbola derived from the asymptotes?
3. What is the relationship between the auxiliary rectangle and the foci of the hyperbola?
4. How do we determine the direction of the axes for a hyperbola?
5. What is the significance of the distance \( c \) in hyperbolas?

**Tip:** For hyperbolas, knowing the asymptotes gives a lot of information about the hyperbola's orientation and dimensions.

An engineer is working on making a diagram for the cooling tower he was going to construct in the future. Cooling towers can be seen as towers with inward-curving sides, or simply, hyperbola-looking sides. His daughter asks him how high the center of the hyperbolic sides of the tower was, and he answers that the center can be situated at (0,15) in the Cartesian plane. The engineer then says that the asymptotes of the hyperbola are the following: y = ±24/7x + 15, and the area of its auxiliary rectangle is 168 square units. The daughter then finds the foci of the hyperbola. At what points on the Cartesian plane are these foci found?

Learn how to find the foci of a hyperbola with its center at (0,15), asymptotes y = ±24/7x + 15, and an auxiliary rectangle area of 168 square units. This problem covers key analytic geometry concepts, including hyperbolas, their asymptotes, and the calculation of foci. Suitable for high school students in Grades 10-12.

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