# Find an equation that models the path of a satellite if its path is a hyperbola, a = 55,000 km, and c = 81,000 km. Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

**Solution:**

The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the x-axis is given by

[(x - h)^{2}/a^{2}] - [(y - k)^{2}/b^{2}] = 1 --- (1)

Where, the length of the transverse axis is 2a.

The coordinates of the vertices are (h ± a, k)

The length of the conjugate axis is 2b

The coordinates of the co-vertices are (h, k ± b)

The distance between the foci is 2c, where c^{2} = a^{2} + b^{2}

The coordinates of the foci are (h ± c, k)

Given, a = 55,000 km

c = 81,000 km

Centre at origin i.e. (h, k) = (0,0)

We know, c^{2} = a^{2} + b^{2}

(81000)^{2} = (55000)^{2} + b^{2}

b^{2} = (81000)^{2} - (55000)^{2}

b^{2} = 3536000000

Taking square root,

b = √(3536000000)

b = 59465 km

Substituting the values in (1)

[(x - 0)^{2}/(81000)^{2}] - [(y - 0)^{2}/(59465)^{2}] = 1

[x^{2}/6561000000] - [y^{2}/3536000000] = 1

Therefore, the equation of the hyperbola is x^{2}/(81000)^{2 }- y^{2}/(59465)^{2} = 1.

## Find an equation that models the path of a satellite if its path is a hyperbola, a = 55,000 km, and c = 81,000 km. Assume that the center of the hyperbola is the origin and the transverse axis is horizontal.

**Summary:**

The equation that models the path of a satellite if its path is a hyperbola with a = 55,000 km, c = 81,000 km. Assuming that the center of the hyperbola is the origin and the transverse axis is horizontal is x^{2}/6561000000 - y^{2}/3536000000 = 1.

visual curriculum