course mth164 ????s???????????assignment #012012. Equations and Properties of Parabolas, hyperbolas and ellipses
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09:04:43 `q001. The set of points which lie at equal distances from a point (x0, y0) and a line y = c is called a parabola. The line y = c is called the directrix and the point (x0, y0) the focus of the parabola. The vertex of the parabola is the point halfway between the focus and the directrix. What is the vertex of the parabola whose focus is at (7, 4) and whose directrix is y = 2? How far is it from the vertex of this parabola to the focus?
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RESPONSE --> Vertex is (7,2) It is one unit from the vertex to the focus. confidence assessment: 2
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09:04:53 Figure 74 shows the point (7,4), now identified as the focus, and the line y = 2, now identified as the directrix. Recall that in the previous exercise set we located points which were equidistant from this line and point. These points are again shown, along with the curve (a parabola) consisting of all such points. The vertex is the point halfway between the focus and the directrix; this point clearly has coordinate form (7, 3), lying 1 unit lower than the focus and one unit higher than the directrix. We often use a to represent the displacement from vertex to focus, or directrix to vertex.
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RESPONSE --> ok self critique assessment:
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09:07:59 `q002. The equation (y - k) = 1 / (4 a) * ( x - h )^2 describes the parabola with vertex (h, k) and displacement a from vertex to focus. Give the equation of the parabola from the preceding problem whose focus is at (7, 4) and whose directrix is at y = -2.
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RESPONSE --> y-2=1/(8)*(x-7)^2 confidence assessment: 3
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09:08:30 The vertex is at (h, k) = (7, 3) and as seen in the preceding solution we have a = 1. Thus the equation is (y - 3) = 1 / (4 * 1) * (x - 7)^2, which simplifies to y - 3 = 1/4 * (x - 7)^2. Note that this is identical with the solution obtained in the preceding exercise, where we simplified the equation sqrt( (x-7)^2 + (y-4)^2 ) = y - 2, representing the statement that the distance from (7,4) to (x,y) is equal to the distance from (x,y) to the line y = 2. After several steps of simplification we obtained the form (y - 3) = 1/4 ( x - 7)^2.
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RESPONSE --> I missed that a = 1. I was thinking it was the directrix value. self critique assessment:
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09:10:05 `q003. What is the equation of the parabola whose vertex is at (8, 2) and whose focus is at (8, 2)? Sketch this parabola and describe its graph.
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RESPONSE --> Since the vertex and the focus are in the same place I am not sure how to write the equation. confidence assessment:
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09:10:13 Figure 35 illustrates the parabola, whose vertex lies 1 unit higher than the focus and whose directrix must therefore lie 1 unit higher than the vertex (recall that the vertex must be equidistant from focus and directrix). The directrix is therefore the line y = 4. The equation of the parabola is (y - k) = 1 / (4 a) * (x - h)^2, where (h,k) is the coordinate form of the vertex and a is the displacement from vertex to focus. Thus (h, k) = (8, 3) so h = 8 and k = 3, while a is the -1 unit displacement from vertex to focus. The equation of the parabola is therefore y - 3 = 1 / (4 * -1) * (x - 8)^2, or y - 3 = -1/4 * (x-8)^2.
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RESPONSE --> ok self critique assessment:
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09:13:54 `q004. The set of points the sum of whose distances from the two given points (0, c) and (0, -c) is equal to 2b is called an ellipse. Its equation is x^2 / a^2 + y^2 / b^2 = 1, where a^2 = b^2 - c^2. The points (0, -c) and (0, c) are called the foci of the ellipse; each individual point is called a focus. What is the equation of the ellipse with foci at (0, -3) and (0, 3) for which a = 5?
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RESPONSE --> x^2/25 + y^2/34 = 1 confidence assessment: 3
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09:14:54 The coordinates of the foci are (-c, 0) and (c, 0), identified here as (-3,0) and (3,0). So we know that c = 3. Since b = 5 and a^2 = b^2 - c^2 we have a = sqrt(b^2 - c^2) = sqrt(5^2 - 3^2) = sqrt(16) = 4. So the equation of the given ellipse is x^2 / a^2 + y^2 / b^2 = 1, with a = 4 and b = 5, or x^2 / 4^2 + y^2 / 5^2 = 1, also written as x^2/16 + y^2/25 = 1.
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RESPONSE --> ok self critique assessment:
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09:17:34 `q005. What are the x intercepts of the ellipse with foci at (0, -3) and (0, 3) for which a = 5? What are the y intercepts?
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RESPONSE --> equation is x^2/25 + y^2/34 = 1 x--intercepts are (5,0) and (-5,0) y-intercepts are (0,sqrt(34) (0,-sqrt(34)) confidence assessment: 3
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09:18:02 The x intercepts of an equation lie at the points where y = 0. So the x intercepts of the equation x^2/16 + y^2/25 = 1 occur when x^2/16 + 0^2/25 = 1. We easily solve the resulting equation: x^2/16 = 1 so that x^2 = 16 and x = +- sqrt(16) = +-4. In general the equation x^2 / a^2 + y^2 / b^2 = 1 has x intercepts at x = +-a and y intercepts at y = +-b. The y intercepts of an equation lie at the points where x = 0. So the y intercepts of the equation x^2/16 + y^2/25 = 1 occur when 0^2/16 + y^2/25 = 1. We easily solve the resulting equation: y^2/25 = 1 so that y^2 = 25 and y = +- sqrt(25) = +-5. In general the equation x^2 / a^2 + y^2 / b^2 = 1 has x intercepts at x = +-a and y intercepts at y = +-b.
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RESPONSE --> I thought a was 5. self critique assessment:
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09:20:43 `q006. Sketch the rectangle formed by the lines x = -4, x = 4, y = -5, and y = 5. Sketch the ellipse circumscribed by this rectangle (this will be the largest ellipse you can fit into the rectangle). What is the equation of this ellipse? What are the coordinates of its foci?
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RESPONSE --> x^2/16 + y^2/25 = 1 Foci are (0,3) (0,-3) confidence assessment: 3
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09:20:55 The rectangle is shown in Figure 99, bounded by the vertical lines x = 4 and x = -4 and by the horizontal lines y = 5 and y = -5. The largest possible ellipse inside the rectangle is shown in Figure 33, intersecting the rectangle at the points (-4,0), (4,0), (0, -5) and (0, 5). Note that the x intercepts of this ellipse are the same as for the equation x^2 / 16 + y^2 / 25 = 1. The general method for sketching an ellipse of this form is to draw vertical lines through the x intercepts and horizontal lines through the y intercepts to form a rectangle. The ellipse can then be inscribed in the rectangle as was done here.
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RESPONSE --> ok self critique assessment:
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09:23:14 `q007. If an ellipse from any of the the preceding exercises is shifted so that its center lies at (h, k) rather than (0, 0), then x and y in its equation will be replaced by (x-h) and (y-k). The rest of its equation will remain the same. What will be the equation of the ellipse which fits into the rectangle defined by the lines x = -2, x = 6, y = -4 and y = 6? Hints: What is the center of this rectangle? If this rectangle is shifted so that the center moves to (0, 0) what will be the values of a and b, and what therefore will be its equation? So what is its equation in its actual position?
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RESPONSE --> (x-2)^2/16 + (y-1)^2/25 = 1 confidence assessment: 3
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09:23:24 The given rectangle is shown in Figure 13. The center of the rectangle will lie halfway between x = 2 and x = 10, at x = (-2 + 6) / 2 = 2, and halfway between y = -4 and y = 6, at y = (-4 + 6) / 2 = 1. The rectangle is therefore centered at (2, 1). If we shifted the rectangle so that the center was that (0, 0), we would have to shift -2 units in the x direction and -1 unit in the y direction. This would shift the lines x = -2 and x = 6 to x = -4 and x = 4, and the lines y = -4 and y = 6 to y = -5 and y = 5. This rectangle would give us the ellipse centered at the origin having intercepts at (-a, 0) = (-4, 0) and (a, 0) = (4, 0), as well as at (-b, 0) = (-5, 0) and (b, 0) = (5, 0). Thus we have a = 0 and b = 0 and the ellipse would have equation x^2 / a^2 + y^2 / b^2 = x^2 / 16 + y^2 / 25 = 0. Shifting this ellipse back to its original position with center at (h, k) = (2, 1), we replace x by x - h = x - 2 and y by y - h = y - 1 to obtain the equation (x-2)^2 / 16 + (y - 1)^2 / 25 = 1. This ellipse is shown in Figure 38.
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RESPONSE --> ok self critique assessment:
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09:24:54 `q008. The set of points whose distances from the two given points (c, 0) and (-c, 0) differ by 2a is called a hyperbola. Its equation is x^2 / a^2 - y^2 / b^2 = 1, where c^2 = a^2 + b^2. The points (-c, 0) and (c, 0) are called the foci of the hyperbola; each individual point is called a focus. What is the equation of the hyperbola with foci at (-5, 0) and (5, 0) for which a = 3?
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RESPONSE --> x^2/9 - y^2/16 = 1 confidence assessment: 3
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09:25:45 The hyperbola with foci at ( +-c, 0) = (+-5, 0), for which a = 3, has b^2 = c^2 - a^3 = 5^2 - 3^2 = 16, so that b = sqrt(16) = 4. So the equation of this ellipse is x^2 / a^2 - y^2 / b^2 = 1, or x^2 / 4^2 - y^2 / 5^2 = 1, which gives us x^2 / 16 - y^2 / 25 = 1.
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RESPONSE --> I thought a was 3 and b was 4 to use in the equation? self critique assessment:
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09:26:45 `q009. Sketch the rectangle formed by the lines x = -4, x = 4, y = -5, and y = 5, using a scale for which the rectangle is fairly small. Sketch the hyperbola defined by this rectangle as follows: First sketch the diagonals of the rectangle, extending them well beyond the rectangle itself. Start at (4, 0) and sketch a curve that starts parallel to the vertical side of the rectangle, then as it rises curves away from the rectangle and becomes asymptotic to the extended diagonal which moves upward and to the right. Then staring again at (4, 0) move downward, gradually curving away from the vertical to become asymptotic to the extended diagonal which moves downward and to the right. You will have sketched the part of the hyperbola whose vertex is at (4,0) and opens to the right. Then sketch the part of the hyperbola which starts at (-4, 0) and opens to the left. This part will be symmetric with the part that opens to the right.
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RESPONSE --> ok confidence assessment: 3
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09:26:54 The rectangle is shown in Figure 82. The diagonals are included in this figure. The hyperbola will have two branches, one starting at (4, 0) and opening to the right. This hyperbola will approach the diagonals as its asymptotes. The other branch starts at (-4, 0) and opens to the left, again approaching the diagonals as its asymptotes. The hyperbola is shown in figure 85.
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RESPONSE --> ok self critique assessment:
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09:27:20 `q010. How does the shape of the hyperbola you just sketched differ from that of a parabola?
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RESPONSE --> I look like to mirror images of a parabola. Both opening right and left confidence assessment: 3
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09:27:29 The parabolas we have encountered so far in the course open upward or downward, while the hyperbola we see here opens to the right and to the left. This however is not really a difference between parabolas and hyperbolas, since parabolas can also open to the right or to the left (but not both on the same parabola). One difference is that a hyperbola has two branches and a parabola does not. Also the hyperbola to approaches asymptotes whereas a parabola never approaches any fixed line. A parabola which opens upward or downward keeps getting steeper and steeper, without limit.
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RESPONSE --> ok self critique assessment:
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09:28:37 `q011. What are the equations of the diagonals of the hyperbola you sketched? How are the equations of the diagonals related to the equation of the hyperbola?
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RESPONSE --> y=5/4x y=-5/4x They form the asymptotes of the hyperbola confidence assessment: 3
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09:28:45 The diagonals both pass through the origin. One has a slope of 5/4 and the other a slope of -5/4, so the equations of the diagonals are y = 5/4 x and y = -5/4 x. The equation of the hyperbola is x^2 / 16 - y^2 / 25 = 1. Solving for y we have y = +-sqrt [ 25 ( x^2 / 16 + 1) ]. For very large x and y, the 1 on the right-hand side is not very significant and we therefore effectively have y = +-sqrt[ 25 x^2 / 16 ](good approximation for large x). sqrt(25 x^2 / 16) = 5 x / 4 so for large x our approximation becomes y = +- 5 x / 4, which is identical to the equations of the asymptotes.
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RESPONSE --> ok self critique assessment:
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09:29:36 `q012. Sketch the rectangle formed by the lines x = -4, x = 4, y = -5, and y = 5, using a scale for which the rectangle is fairly small. Sketch the hyperbola defined by this rectangle as follows: First sketch the diagonals of the rectangle, extending them well beyond the rectangle itself. Start at (5, 0) and sketch a curve that starts parallel to the horizontal side of the rectangle, then as it moves to the right curves away from the rectangle and becomes asymptotic to the extended diagonal which moves upward and to the right. Then staring again at (5, 0) move to the left, gradually curving away from the horizontal to become asymptotic to the extended diagonal which moves upward and to the left. You will have sketch the part of the hyperbola whose vertex is at (5,0) and opens upward. Then sketch the part of the hyperbola which starts at (-5, 0) and opens downward.
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RESPONSE --> ok confidence assessment:
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09:30:02 Figure 51 shows the hyperbola as described. It is not completely clear from the sketch how the hyperbola approaches its asymptotes. The approach to the asymptote is illustrated for the lower branch in Figure 6.
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RESPONSE --> ok self critique assessment:
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09:31:32 `q013. The hyperbola you just sketched, which opens upward, has equation y^2 / b^2 - x^2 / a^2 = 1 and its foci lie at (0, c) and (0, -c). What is the equation of this hyperbola?
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RESPONSE --> y^2/25 - x^2/16 = 1 confidence assessment: 2
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09:31:40 The foci of the parabola were given as (0, -c) = (0, -3) and (0, c) = (0, 3), so c = 3. The value of b is that of the y intercepts so b = 5. It follows that a^2 = b^2 - c^2 = 5^2 - 3^2 = 16, so that a = sqrt(16) = 4. The equation is therefore y^2 / b^2 - x^2 / a^2 = 1, or y^2 / 5^2 + x^2 / 4^2 = 1, or y^2 / 25 - x^2 / 16 = 1.
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RESPONSE --> ok self critique assessment:
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09:47:04 `q014. The hyperbola y^2 / b^2 - x^2 / a^2 = 1 has points on the 'top' and 'bottom', not on the 'right' and 'left' of the defining rectangle. How can you tell this from the equation? How can you tell that the equation x^2 / a^2 - y^2 / b^2 = 1 must have points at the 'right' and 'left' but not at the 'top' and 'bottom' of the defining rectangle?
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RESPONSE --> Because the y variable is first Because the x variable is first confidence assessment: 3
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09:47:10 For the equation y^2 / b^2 - x^2 / a^2 = 1, the y-intercept will lie at the x = 0 points, where y^2 / b^2 - 0^2 / a^2 = 1. Simplifying this equation we have y^2 / b^2 = 1, or y^2 = b^2 so that y = +-b. However if we attempt to find the x intercepts we substitute y = 0 and obtain -x^2 / a^2 = 1. This equation has no solution since x^2 and a^2 are both positive, making - x^2 / a^2 negative. Since the negative quantity cannot equal 1, there is no x intercept--so, there is no graph point on the right and left sides of the defining rectangle. For the equation x^2 / a^2 - y^2 / b^2 = 1 we see by letting y = 0 that the x intercepts are x = +-a, but if we let x = 0 we get the equation - y^2 / b^2 = 1, with no solution. We therefore have x intercepts but not y intercepts--i.e., points on the right and left sides of the rectangle but not on top and bottom.
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RESPONSE --> ok self critique assessment:
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