Question 1185983
Here's how to solve this problem:

**A. Find the coordinates of the focus and the equations of the asymptotes:**

1.  **Hyperbola Orientation:** Since the jet is flying north and the vertex is behind it, the hyperbola opens north (upwards in our coordinate system). This means the hyperbola's transverse axis is vertical.

2.  **Center:** The center of the hyperbola is halfway between the jet (origin) and the vertex. Since the vertex is 15 km behind (south of) the jet, the center is 15/2 = 7.5 km south of the jet.  Therefore, the center is at (0, -7.5).

3.  **Distance from Center to Vertex (a):** The distance from the center to the vertex is 7.5 km.  So, a = 7.5.

4.  **Distance from Center to Focus (c):** We're given that the eccentricity (e) is 1.5.  Eccentricity is defined as e = c/a.  Therefore:

    1.5 = c / 7.5
    c = 1.5 * 7.5 = 11.25

5.  **Focus:** The focus is 'c' units from the center along the transverse axis. Since the hyperbola opens upwards, the focus is 11.25 km south of the jet. Therefore, the focus is at (0, -11.25).

6.  **Equations of the Asymptotes:** For a hyperbola with a vertical transverse axis, the equations of the asymptotes are:

    y - k = ±(a/b)(x - h)

    where (h, k) is the center, 'a' is the distance from the center to the vertex, and 'b' is related to 'a' and 'c' by the equation c² = a² + b².

    First, we find b:
    11.25² = 7.5² + b²
    b² = 126.5625 - 56.25
    b² = 70.3125
    b = √70.3125 ≈ 8.385

    Now, substitute the values into the asymptote equations:
    y + 7.5 = ±(7.5/8.385)(x - 0)
    y ≈ ±0.894x - 7.5

**B. Equation of the Hyperbola:**

The standard form equation for a hyperbola with a vertical transverse axis and center (h, k) is:

((y - k)² / a²) - ((x - h)² / b²) = 1

Substituting the values we found:

((y + 7.5)² / 7.5²) - (x² / 70.3125) = 1

((y + 7.5)² / 56.25) - (x² / 70.3125) = 1

**C. Time until John hears the sonic boom:**

1.  **Equation of the Hyperbola's Leading Edge:**  The leading edge of the sonic boom is represented by the *right* branch of the hyperbola. Since John is at (7,10), we can plug in x=7 and solve for the y coordinate of the hyperbola's leading edge.

((10 + 7.5)² / 56.25) - (7² / 70.3125) = 1
(17.5²/56.25) - (49/70.3125) = 1
5.44 - 0.697 = 4.743
y = 10.743

2. **Distance Traveled by the Sonic Boom:** The leading edge of the hyperbola is at approximately y = 10.743 km when x=7.  The hyperbola is moving north at 500m/s. The distance that the sonic boom has traveled is approximately 10.743 km + 7.5 km = 18.243km.

3.  **Time:** Time = Distance / Speed = 18243 m / 500 m/s ≈ 36.5 seconds

**Answers:**

*   **A.** Focus: (0, -11.25). Asymptotes: y ≈ ±0.894x - 7.5
*   **B.** ((y + 7.5)² / 56.25) - (x² / 70.3125) = 1
*   **C.** Time ≈ 36.5 seconds