A satellite orbits Earth in an elliptical path with Earth at one of the foci. If the major axis of the ellipse is 20,000 km and the minor axis is 16,000 km, find the distance from the center of the ellipse to each focus. - 4.6 Conic Sections - Review - Use this assignment in your class!

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Given the equation of a conic section

4 x^{2} + 9 y^{2} = 36

, identify the type of conic section and its key features.

Given the equation of a parabola

y^{2} = 4 a x

, sketch the graph and identify the vertex and focus of the parabola.

Given the equation of an ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

where

a > b

, sketch the graph and identify the vertices and foci.

Given the equation of a hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

, sketch the graph and identify the vertices and asymptotes.

Given the equation of a circle

(x - h)^{2} + (y - k)^{2} = r^{2}

, sketch the graph and identify the center and radius.

Given the equation of a hyperbola

\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1

, sketch the graph and identify the vertices and foci.

A parabolic reflector is used in a satellite dish to focus signals onto a receiver located at the focus. If the equation of the parabola is

y^{2} = 16 x

, find the coordinates of the focus.

In an engineering project, a hyperbolic cooling tower is designed with the equation

\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1

. Determine the coordinates of the vertices and the foci of the hyperbola.

In physics, the path of a comet around the sun is often modeled as a conic section. If a comet follows a parabolic trajectory given by the equation

x^{2} = 4 p y

, where

p

is the distance from the vertex to the focus, and the vertex is at the origin, find the value of

p

if the focus is at (0, 3).

An engineer is designing a bridge with a cross-section in the shape of a hyperbola. The equation of the hyperbola is

\frac{y^{2}}{9} - \frac{x^{2}}{16} = 1

. Determine the asymptotes of the hyperbola.

The distance from the center of the ellipse to each focus, denoted as

c

, can be found using the relationship

c^{2} = a^{2} - b^{2}

, where

a

is the semi-major axis and

b

is the semi-minor axis. Here,

a = \frac{20, 000}{2} = 10, 000

km and

b = \frac{16, 000}{2} = 8, 000

km. Thus,

c^{2} = 10, 000^{2} - 8, 000^{2} = 100, 000, 000 - 64, 000, 000 = 36, 000, 000

. Therefore,

c = \sqrt{36, 000, 000} = 6, 000

km. The distance from the center to each focus is 6,000 km.

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