# A satellite orbits in an elliptical path such that the Earth's center is one of the foci of the ellipse. The distance from

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## Answers

[tex]\frac{x^2}{8200^2} +\frac{y^2}{7101.4^2}=1[/tex]

Step-by-step explanation:

The equation of the ellipse with centre at origin and major axis along the x-axis is:

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

Where 2a is the length of the major axis and 2b is the length of the minor axis. The following equation shows the relationship between the foci (c), major axis and minor axis:

a² = c² + b²

Since the earth center is one of the foci of the ellipse with distance from the center of the orbit to the Earth's center is 4,100 miles. Therefore the foci is at (4100, 0).

Hence c = 4100

Also, The foci are halfway between the center and the major axis vertices. Therefore the distance between the center and the major axis vertices (a) is twice the distance from the center of the orbit to the Earth's center.

a = 2c = 2(4100) = 8200 miles

We can get the distance from the center to the minor axis vertices (b), using:

a² = c² + b²

8200² = 4100² + b²

b² = 8200² - 4100²

b² = 50430000

b=√50430000

b = 7101.4 miles

The ellipse equation is:

[tex]\frac{x^2}{8200^2} +\frac{y^2}{7101.4^2}=1[/tex]

[tex]\frac{x^{2} }{67,240,000}+\frac{y^{2} }{50,430,000} =1[/tex]

Step-by-step explanation:

2 x t divided by n

step-by-step explanation:

answer: d. $20

step-by-step explanation:

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