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10 points to the person who can help me with 4 precalculus questions?!?!?! please!?

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1. Write an equation of an ellipse in standard form with the center at the origin and with the given characteristics. vertex at (–5, 0) and co-vertex at (0, 4)

2. A hyperbola has vertices (±5, 0) and one focus (6, 0). What is the standard-form equation of the hyperbola?


3. What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 – 4y^2 = 64?


4.. 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.

asked Mar 28, 2013 in PRECALCULUS by andrew Scholar

4 Answers

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The standard form of an ellipse equation  with the center at the origin :x2 / a2 + y2 / b2 = 1

Where a is the major axis and b is the minor axis

Substitute the vertex (-5,0) in the equation of an ellipse : x2 / a2 + y2 / b2 = 1

(-5)2 /a2 + 02 / b2 = 1graph the ellipse x^2/25 + y^2/16 =1

25 /a2 + 0 = 1

25 /a2 = 1

Multiply each side by a2

25 = a2

a2 = 25

Substitute covertex (0,4) in an ellipse equation : x2 / a2 + y2 / b2 = 1

02 / a2 + 42 / b2 = 1

0 + 16 / b2 = 1

16 / b2 = 1

Multiply each side by b2

16 = b2

b2 = 16

Substitute a2 = 25 and b2 = 16 in an ellipse equation : x2 / a2 + y2 / b2 = 1

There fore The equation of an ellipse at the origin : x2 / 25 + y2 / 16 = 1.

 

 

answered Mar 28, 2013 by diane Scholar
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2.

A vertices of  hyperbola (-5,0) and (5,0)

The center of the hyperbola is two vertices mid point

There fore mid point ( center) is (5 -5 / 2, 0 + 0 / 2) = (0 , 0)

A focus on hyperbola (6, 0)

The center (h , k)  Length of the transverse axis 2a and length of conjugate axis 2b then

The equation of the hyperbola = (x - h)2 / a2 - (y - k)2 / b2 = 1

Length of the transverse axis 2a = 5 - (-5)

2a = 5 + 5

2a = 10

Divide each side by 2

a = 5

Take square each side a2 = 52

a2 = 25

c  is the distance from  center to focus  on horizontal transverse = 6 - 0

c = 6

Take square to each side

c2 = 36

But c2 = a2 + b2

Substitute a2 = 25 and c2 = 36 in the c2  = a2 + b2

36 =25 + b2

Subtract 25 from each side

11 = b2

b2 = 11

Take square root to each side

b = sqrt(11) , -sqrt(11)

The standard form of the hyperbola is (x - h)2 / a2 - (y - k)2 / b2 = 1

Substitute h =0, k = 0, a2 =25  and b2 = 11 in the standard equation form of hyperbola

(x -0)2 / 25 - (y -0)2 / 11 = 1

x2 / 25 - y2 / 11 = 1

The standard equation form of hyperbola : x2 / 25 - y2 / 11 = 1.

answered Mar 28, 2013 by diane Scholar
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3) The hyperbola equation is image

The standard form for hyperbola is in a form = 1, So divide both sides of equation by 64 to set it equal to 1.

image

image

image

Compare it to standard form of hyperbola image

"a " is the number in the denominator of the positive term

If the x -term is positive, then the hyperbola is horizontal

a = semi-transverse axis = 2

b = semi-conjugate axis = 4

center: (h, k ) = (0,0)

Vertices: (h + a, k ), (h - a, k )

= (0+ 2, 0) , (0 - 2, 0)

Vertices of the hyperbola  (2,0) , (-2,0)

c  = distance from the center to each focus along the transverse axis

image

image

Foci: (h + c, k ), (h - c, k )

image

Foci of the hyperbola image

The equations for asymptotes are 

image

image

image

Asymptotes of hyperbola are  image.

answered Jun 6, 2014 by david Expert
0 votes

4) A satellite if its path is a hyperbola,

Transverse axis is horizontal then standard form of hyperbola is image

Center of hyperbola is origin, So (h ,k ) = (0, 0).

c  = 81000km, a  = 55000km

c  = distance from the center to each focus along the transverse axis.

image

image

image

image

image

image

Substitute the values of (h ,k ) and a,b in horizontal hyperbola equation.

image

So the equation is image.

answered Jun 7, 2014 by david Expert

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