#$&* course Mth 164 03/25 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (y-k)^2=4a(x-h) (y + 2)^2 =8(x - 4) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: GOOD STUDENT 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: query problem 9.2.42 y^2+12y = -x+1.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2+12y = (y^2 + 12 y + 6^2) - 6^2 (y^2 + 12 y + 36) - 36 (y+6)^2 - 36. (y+6)^2 - 36 = - x + 1 (y - (-6))^2 = -x + 37 (y-k)^2 = 4 a ( x - h) -x + 37 = -1(x - 37) = 4 ( -1/4) ( x - h) (y-k)^2 = 4 (-1/4) ( x - 37) Vertex (-6, 37) Focus -.25 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Completing the square on the left-hand side we have y^2+12y = (y^2 + 12 y + 6^2) - 6^2 or (y^2 + 12 y + 36) - 36, which we put in the form (y+6)^2 - 36. The equation becomes (y+6)^2 - 36 = - x + 1, which we rearrange to get (y - (-6))^2 = -x + 37. The form we need is (y-k)^2 = 4 a ( x - h). The right-hand side is -x + 37 = -1(x - 37) = 4 ( -1/4) ( x - h), so the equation is (y-k)^2 = 4 (-1/4) ( x - 37). The vertex is therefore at (-6, 37). The focus lies at displacement -1/4 = -.25 in the x direction from the vertex, at (36.75, -6). The directrix is at x = 37 - (-.25) = 37.25. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The directrix = 37.25 ------------------------------------------------ Self-critique Rating: **** query problem 9.2.48 Vertex (1,-1), y-intercept (0,1), opens upward.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (y + 1) = 2 ( x - 1)^2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The parabola has form (y-h) = 1 / (4 a) * ( x - k )^2 with vertex (h, k) and vertex-to-focus displacement a. The vertex is (1,-1) so we have (y - -1) = 1 / (4 a) * (x - 1) or (y + 1) = 1 / (4 a) * (x - 1)^2. Substituting (0,1) into the equation we have (1 + 1) = 1 / (4a) * ( 0 - 1 )^2, or 2 = 1 / (4a) so that a = 1/8; this makes 1 / (4 a) = 1/(4 * 1/8) = 1 / (1/2) = 2. Thus (y + 1) = 2 ( x - 1)^2. We can verify this by noting that a parabola with vertex at (1,-1) and y-intercept (0,1) is just the graph of y = x^2 stretched by factor 2 to give y = 2 x^2 then shifted 1 unit right and 1 unit down to give (y + 1) = 2 ( x - 1)^2. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: query problem 9.3.18 eqn of ellipse center at (0,0); focus at (0,1); vertex at (0,-2).
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 / 3 + y^2 / 4 = 1 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The center is (0, 0). Since the given focus is on the y axis, the major axis must be on the y axis. • An ellipse with center at the origin has equation x^2 / a^2 + y^2 / b^2 = 1. • When y = 0 we have x^2 / a^2 = 1, so that x^2 = a^2 and therefore x = +- a . • So the ellipse intersects the x axis at (-a, 0) and (a, 0). These are two of the vertices of the ellipse. • When x = 0 we find that y = +- b, so the y intercepts are at (-b, 0) and (b, 0). These are the other two vertices. The vertex (0, -2) is given, which tells us that b = 2. • Since the major axis is in the y direction, we have a^2 = b^2 - c^2, where c is the distance of a focus from the center. We know that b = 2 and c = 1 so we have • a^2 = b^2 - c^2 = 2^2 - 1^2 = 3. The correct equation is therefore • x^2 / 3 + y^2 / 4 = 1. **
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15:29:06 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** query problem 9.3.40 find the center, foci and vertices of the ellipse given by the equation 9x^2+y^2-18x=0
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 + 9 ( x^2 - 2x) = 0 y^2 + 9 ( x^2 - 2 x + 1 - 1) = 0 y^2 + 9 ( (x-1)^2 - 1) = 0 y^2 + 9(x-1)^2 = 9 y^2 / 3^2 + (x-1)^2 / 1^2 = 1 Vertices (2, 0), (0, 0), (1, -3) and (1, 3) 9(-1)^2 + 3^2 - 18 (1) 9 + 9 - 18 = 0 9(0)^2 + 0^2 - 18 (1) 0 + 0 - 18 = -18 9(2)^2 + 0^2 - 18 (0) 18 + 0 - 18 = 0 9(1)^2 + 3^2 - 18 (3) 9 + 9 - 18 = 0 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 9 x^2 + y^2 - 18 x = 0. Factoring the 9 out of the x terms we have y^2 + 9 ( x^2 - 2x) = 0. Completing the square in the parentheses we get y^2 + 9 ( x^2 - 2 x + 1 - 1) = 0, or y^2 + 9 ( (x-1)^2 - 1) = 0, or y^2 + 9(x-1)^2 - 9 = 0 or y^2 + 9(x-1)^2 = 9. Dividing both sides by 9 we have y^2 / 9 + (x-1)^2 / 1 = 1, or y^2 / 3^2 + (x-1)^2 / 1^2 = 1. The semiaxis in the x direction is 1, the semiaxis in the y direction is 3, the center is (1,0). So the vertices lie 1 unit right and left of the center, and 3 units above and below the center. This puts the vertices at (2, 0) and (0, 0), and at (1, -3) and (1, 3). The coordinates of any of the vertices can be plugged into 9x^2+y^2-18x=0 to obtain an identity. For example, (1, 3) gives us 9(-1)^2 + 3^2 - 18 (1) = 9 + 9 - 18 = 0. Plugging in the coordinates of all 4 vertices verifies our results. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: 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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15:57:24 153.5 million miles is the aphelian. the equation would be (x^2/128.5) +(y^2/153.5)=1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aphelion a + c = 155.5 million miles with b^2 = a^2 - c^2 = 142^2 - 13.5^2 = 19982 b = sqrt(19982) = 141.36 x^2 / a^2 + y^2 / b^2 = 1, i.e., x^2 / (142)^2 - y^2 / (141.36)^2 = 1. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!