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

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.

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 = 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.

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.

3) The hyperbola equation is

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

Compare it to standard form of hyperbola

"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

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

Foci of the hyperbola

The equations for asymptotes are

Asymptotes of hyperbola are  .

4) A satellite if its path is a hyperbola,

Transverse axis is horizontal then standard form of hyperbola is

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.

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

So the equation is .