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.

Answer 1

1. The standard form equation of a hyperbola with vertices (±4,0) and one focus (5,0) is:

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

The distance between the center (h,k) and each vertex is a = 4, so a = 4. The distance between the center and the focus is c = 5, so c = 5. By using the relationship a^2 + b^2 = c^2, we can solve for b:

4^2 + b^2 = 5^2
16 + b^2 = 25
b^2 = 9
b = 3

Therefore, the equation of the hyperbola is:

(x - 0)^2 / 4^2 - (y - 0)^2 / 3^2 = 1
x^2 / 16 - y^2 / 9 = 1

2. To find the vertices, foci, and asymptotes of the hyperbola with the equation 20x^2 - 25y^2 = 100, we first need to rewrite it in standard form:

x^2 / 5 - y^2 / 4 = 1

The vertices are at ±(√5, 0) and ±(0, √4) which simplifies to (±√5, 0) and (0, ±2).

The foci can be found using the formula c = √(a^2 + b^2), which gives us c = √(5^2 + 4^2) = √41. Therefore, the foci are at ±(√41, 0).

The asymptotes of the hyperbola are given by y = ±b/a * x, where a = √5 and b = √4. So, the asymptotes are y = ±2/√5 * x.

3. The general form equation of a hyperbola centered at the origin with a horizontal transverse axis is:

x^2 / a^2 - y^2 / b^2 = 1

Given 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 = 3,016,000,000
b = √3,016,000,000

Therefore, the equation that models the path of the satellite is:

x^2 / 45,000^2 - y^2 / 3,016,000,000 = 1