Question
1. A hyperbola has vertices (+4,0) and one focus (5,0). What is the standard-form equation of the hyperbola?
2. State the vertices, foci, and asymptotes of the hyperbola with the equation
20x^2 - 25y2^ =100
3. Find an equation that models the path of a satellite if its path is a hyperbola, a = 45,000 km, and c = 71,000 km.
Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.
2. State the vertices, foci, and asymptotes of the hyperbola with the equation
20x^2 - 25y2^ =100
3. Find an equation that models the path of a satellite if its path is a hyperbola, a = 45,000 km, and c = 71,000 km.
Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.
Answers
GPT 3.5
1. Since the hyperbola has vertices at (+4,0) and one focus at (5,0), we can determine that the center of the hyperbola is at (4,0). The distance from the center to a focus is the value of "c" in the standard form equation of a hyperbola. In this case, c = 1.
The standard form equation of a hyperbola with center (h,k), vertices (h ± a, k), and foci (h ± c, k) is given by:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
Plugging in the values we have:
Center: (4,0)
Vertices: (4 ± a, 0) = (4 ± 1, 0) = (3,0) and (5,0)
Foci: (4 ± c, 0) = (4 ± 1, 0) = (3,0) and (5,0)
Since the transverse axis is on the x-axis, a = 1. Therefore, the equation of the hyperbola is:
(x-4)^2/1 - y^2/b^2 = 1
(x-4)^2 - y^2/b^2 = 1
2. To find the vertices, foci, and asymptotes of the hyperbola with the equation 20x^2 - 25y^2 = 100, we need to rewrite the equation in standard form. Divide both sides by 100 to get:
x^2/5 - y^2/4 = 1
Comparing this equation to the standard form, we can see that a = √5 and b = √4.
Vertices: (±√5, 0)
Foci: c^2 = a^2 + b^2 = 5 + 4 = 9, so c = 3
Foci: (±3, 0)
The asymptotes of the hyperbola are y = ±(b/a)x, so in this case, they are y = ±(√4/√5)x or y = ±(2/√5)x.
3. The equation of a hyperbola with center at the origin, transverse axis along the x-axis, a = 45,000 km, and c = 71,000 km is given by:
x^2/a^2 - y^2/b^2 = 1
Since the center is at the origin, (h,k) = (0,0), and a = 45,000 km and c = 71,000 km. We can find b using the relationship a^2 = b^2 + c^2:
(45,000)^2 = b^2 + (71,000)^2
2,025,000,000 = b^2 + 5,041,000,000
b^2 = 5,041,000,000 - 2,025,000,000
b^2 = 3,016,000,000
b = √3,016,000,000 ≈ 54,937 km
Therefore, the equation of the hyperbola is:
x^2/(45,000)^2 - y^2/(54,937)^2 = 1
The standard form equation of a hyperbola with center (h,k), vertices (h ± a, k), and foci (h ± c, k) is given by:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
Plugging in the values we have:
Center: (4,0)
Vertices: (4 ± a, 0) = (4 ± 1, 0) = (3,0) and (5,0)
Foci: (4 ± c, 0) = (4 ± 1, 0) = (3,0) and (5,0)
Since the transverse axis is on the x-axis, a = 1. Therefore, the equation of the hyperbola is:
(x-4)^2/1 - y^2/b^2 = 1
(x-4)^2 - y^2/b^2 = 1
2. To find the vertices, foci, and asymptotes of the hyperbola with the equation 20x^2 - 25y^2 = 100, we need to rewrite the equation in standard form. Divide both sides by 100 to get:
x^2/5 - y^2/4 = 1
Comparing this equation to the standard form, we can see that a = √5 and b = √4.
Vertices: (±√5, 0)
Foci: c^2 = a^2 + b^2 = 5 + 4 = 9, so c = 3
Foci: (±3, 0)
The asymptotes of the hyperbola are y = ±(b/a)x, so in this case, they are y = ±(√4/√5)x or y = ±(2/√5)x.
3. The equation of a hyperbola with center at the origin, transverse axis along the x-axis, a = 45,000 km, and c = 71,000 km is given by:
x^2/a^2 - y^2/b^2 = 1
Since the center is at the origin, (h,k) = (0,0), and a = 45,000 km and c = 71,000 km. We can find b using the relationship a^2 = b^2 + c^2:
(45,000)^2 = b^2 + (71,000)^2
2,025,000,000 = b^2 + 5,041,000,000
b^2 = 5,041,000,000 - 2,025,000,000
b^2 = 3,016,000,000
b = √3,016,000,000 ≈ 54,937 km
Therefore, the equation of the hyperbola is:
x^2/(45,000)^2 - y^2/(54,937)^2 = 1
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