Final assignment 12

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This is correct; you don't show how you got the result.

Presumably you used a method similar to that used by another student, who submitted the following solution:

we can first draw the graph and plot the points for the vertex and the

focus.

since the focus is to the right of the vertex we know that the

parabola opens to the right.

we can then use the equation (y-k)^2=4a(x-h) with (h, k) = (4, -2) being the coordinates of the vertex.

the displacement from vertex (4, -2) to focus (6, -2) is +2 units so that a = 2.

The equation is therefore (y - (-2) ) ^2 = 4 * 2 (x - 4) or

(y+2)^2 = 8 ( x - 4).

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Good. Compare your details with the following;

** Completing the square on the left-hand side we have

y^2+12y = (y^2 + 12 y + 6^2) - 6^2 or

(y^2 + 12 y + 36) - 36, which we put in the form

(y+6)^2 - 36.

The equation becomes

(y+6)^2 - 36 = - x + 1, which we rearrange to get

(y - (-6))^2 = -x + 37.

The form we need is

(y-k)^2 = 4 a ( x - h).

The right-hand side is -x + 37 = -1(x - 37) = 4 ( -1/4) ( x - h), so the equation is

(y-k)^2 = 4 (-1/4) ( x - 37).

The vertex is therefore at (37, -6). The focus lies at displacement -1/4 = -.25 in the x direction from the vertex, at (36.75, -6). The directrix is at x = 37 - (-.25) = 37.25. **

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10:39:06

`questionNumber 120000

How did you obtain your results?

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RESPONSE -->

i completed the square first and then used the formulas

V(h,k)

F(h+a,k)

D: x=h-a

With h, k, and a from the formula

(y-k)^2=4a(x-h)

confidence assessment: 3

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10:40:26

`questionNumber 120000

query problem 9.2.48 Vertex (1,-1), y-intercept (0,1), opens upward.

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RESPONSE -->

ok

confidence assessment:

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10:40:46

`questionNumber 120000

What is the equation of the given parabola?

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RESPONSE -->

(x-1)^2=.5(y+1)

confidence assessment: 3

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10:41:36

`questionNumber 120000

How did you obtain your result?

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RESPONSE -->

I started with the formula

(x-1)^2=4a(y+1) putting in the h,k from the vertex

then substituting in the point (0,1) I found a to be 1/8 yielding the final equation

confidence assessment: 3

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10:42:20

`questionNumber 120000

query problem 9.3.18 eqn of ellipse center at (0,0); focus at (0,1); vertex at (0,-2).

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RESPONSE -->

ok

confidence assessment:

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10:42:59

`questionNumber 120000

What is the equation of the given ellipse?

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RESPONSE -->

x^2/3+y^2/4=1

confidence assessment: 3

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10:49:08

`questionNumber 120000

How did you obtain your results?

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RESPONSE -->

from the foci c = 1 and from the vertex a = -2

find b using b^2=a^2-c^2

Plug into formula x^2/b^2+y^2/a^2=1

confidence assessment: 3

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10:51:24

`questionNumber 120000

query problem 9.3.40 find the center, foci and vertices of the ellipse given by the equation 9x^2+y^2-18x=0

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RESPONSE -->

ok

confidence assessment:

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10:55:35

`questionNumber 120000

what are the center, foci and vertices of the ellipse?

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RESPONSE -->

Center (1,0)

Foci (1,2sqrt(2) and (1,-2sqrt(2))

Vertices: (1,3) and (1,-3)

confidence assessment: 3

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10:58:14

`questionNumber 120000

How are these quantities obtained?

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RESPONSE -->

I first completed the square with the x and then divided by 9 and got

(x-1)^2 + y^2/9 = 1

so h=1, k=0, a=3, b=1, c=2sqrt(2)

Using the formulas center(h,k)

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

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

confidence assessment: 3

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10:59:35

`questionNumber 120000

query problem 9.3.64 The mean distance of Mars from the sun is 142 million miles and its perihelion is 128.5 million miles. What is the aphelian??Note that full problem statement is the same in both the 5th and 6th editions.

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RESPONSE -->

ok

confidence assessment:

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11:05:08

`questionNumber 120000

What is any equation for the orbit of Mars around the sun?

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RESPONSE -->

The aphelion is 155.5 million miles

x^2/20164 = y^2/19982=1

confidence assessment: 2

Your solution is consistent with the following.

The aphelion is the greatest distance from the Sun and the perihelion is the closest approach. The Sun is at one of the foci of the ellipse.

The periohelion and aphelion are along the same axis, with the foci also along this axis. The Sun is at one of the foci.

To see these it's a good idea to draw a sketch of an ellipse. Let the major axis be the x axis (it could as well be the y axis, or in fact any straight line; but we lose nothing by assuming the axis to be the x axis) with the center of the ellipse at the origin. The ellipse goes through the x axis at (-a, 0) and (a, 0).

The Sun is at one of the foci; let the Sun be at the point (-c, 0).

From the point (-a, 0) to (-c, 0) is the perihelion, and from (-c, 0) to (a, 0) is the aphelion. Thus the perihelion is a - c and the aphelion is a + c.

So a is halfway between the perihelion and the aphelion--i.e., a is the mean distance of the Earth from the Sun.

a - c is the thus 128.5 million miles, a is 142 million miles and c is the difference 13.5 million miles.

The aphelion is therefore a + c = 155.5 million miles.

b is the semi-minor axis, with b^2 = a^2 - c^2 = 142^2 - 13.5^2 = 19982, so that b = sqrt(19982) = 141.36, with distances measured in millions of miles.

So an equation of the ellipse could be

x^2 / a^2 + y^2 / b^2 = 1, i.e.,

x^2 / (142)^2 - y^2 / (141.36)^2 = 1.

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Good responses. See my notes and let me know if you have questions. &#