Solve the problem.
-A satellite following the hyperbolic path shown in the picture turns rapidly at (0, 5)and then moves closerand closer to the line as it gets farther from the tracking station at the origin. Find the equation thatdescribes the path of the rocket if the center of the hyperbola is at (0, 0).

Question

Solve the problem.
-A satellite following the hyperbolic path shown in the picture turns rapidly at (0, 5)and then moves closerand closer to the line  as it gets farther from the tracking station at the origin. Find the equation thatdescribes the path of the rocket if the center of the hyperbola is at (0, 0).

Quizplus · Accepted Answer

Since the hyperbola turns rapidly at (0,5) and gets closer to the line as it moves farther away from the origin, this means that the transverse axis is vertical and the foci are above and below the center. Additionally, the line of symmetry of the hyperbola is the line y=5.

Using the distance formula and the fact that the center is at (0,0), we can find the distance between the center and one of the foci:

c = sqrt(a^2 + b^2) where a is the distance from the center to a vertex and b is the distance from the center to a point on the hyperbola where it intersects the line y=5.

By looking at the graph, we can estimate that a=5 and that one of the points on the hyperbola where it intersects y=5 is (6,5). Therefore, we have:

c = sqrt(5^2 + 1^2) = sqrt(26)

The equation of the hyperbola in standard form is:

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

Since the transverse axis is vertical, the equation becomes:

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

Plugging in the values we know, we have:

a = 5
c = sqrt(26)
h = 0
k = 0

To find b, we can use the relationship:

b^2 = c^2 - a^2

b^2 = 26 - 25 = 1

b = 1

Plugging all these values into the equation, we get:

y^2 / 25 - x^2 / 1 = 1

Simplifying:

y^2 - 25x^2 = 25

Therefore, the equation that describes the path of the rocket is:

Answer: A) 11ebbed0_ca97_9a89_8acb_cd2d02b14ca6_TB8253_11

Solve the Problem. -A Satellite Following the Hyperbolic Path Shown in the Picture