#$&* course Mth 164 03/11 9:57 011. Conic sections
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Given Solution: `aThe distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus 7 - 3 = 4 and 2 - 5 = -3, and the hypotenuse is sqrt(4^2 + (-3)^2) = sqrt(25) = 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. What is the distance from (7, 2) to (x, y)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: D = sqrt((x - x)^2 + (y - y)^2) D = sqrt((x - 7)^2 + (y - 2)^2) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x - 7 and y - 2, and the hypotenuse is sqrt((x-7)^2 + (y-2)^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The legs of the triangle are x-7 and y-2. The hypotenuse is sqrt((x-7)^2 + (y-2)^2) ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q003. What is the distance from (x1, y1) to (x, y)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The legs of the triangle are (x - x1) and (y - y1). The hypotenuse is sqrt((x - x1) + (y - y1)). confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x - x1 and y - y1, and the hypotenuse is sqrt((x-x1)^2 + (y-y1)^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. What is the distance from (x1, y1) to (x2, y2)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The legs of the triangle are (x2 - x1) and (y2 - y1). The hypotenuse is sqrt((x2 - x1)^2 + (y2 - y1)^2). confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance is the hypotenuse of a right triangle with vertices at the two points, whose legs are parallel to the x and y axes and whose hypotenuse runs from one point to the other. The legs of the triangle are thus x2 - x1 and y2 - y1, and the hypotenuse is sqrt((x2-x1)^2 + (yy-21)^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Write as an equation: The distance from (7, 2) to (x, y) is 9. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9 = sqrt(x-7)^2 + (y - 2)^) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), so the statement says that sqrt((x-7)^2 + (y-2)^2) = 9. Note that both sides of this equation could be squared to get (x-7)^2 + (y-2)^2 = 81. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Bothe sides can be squared to get rid of the sqrt and make the 9 equal to 81. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. Write as a system of two equations: The distance from (7, 2) to (x, y) is 9 and distance from (4, 1) to (x, y) is 10. Solve the system for x and y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 9 = sqrt(x - 7)^2 + (y - 2)^2) 10 = sqrt(x - 4)^2 + (y - 1)^2) 81 = (x - 7)^2 + (y - 2)^2) 100 = (x - 4)^2 + (y - 1)^2) x^2 - 14 x + 49 + y^2 - 4 y + 4 = 81 x^2 - 8 x + 16 + y^2 - 2 y + 1 = 100 x^2 - 14 x + y^2 - 4 y = 28 x^2 - 8 x + y^2 - 2 y = 83 -6x - 2 y = -55 -2y = 6x - 55 y = -3x + 55/2 x^2 - 14 x + 49 + (-3x + 55/2)^2 - 4 ( -3x + 55/2 ) + 4 = 81 10·x^2 - 167·x + 701.25 = 83 x = 11.16, 5.54. y^2 - 2·y + 35.2656 = 83 y = 7.98, -5.98 y^2 - 2·y - 34071/2500 = 83 y = 10.88, -8.88. (11.16, 7.98), (11.16,-5.98), (5.54, 10.88), and (5.54, -8.88) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), so the first statement says that sqrt((x-7)^2 + (y-2)^2) = 9. The distance from (4, 1) to (x, y) is sqrt((x-4)^2 + (1-y)^2), so the second statement says that sqrt((x-4)^2 + (1-y)^2) = 10.{} If we square both equations we get (x-7)^2 + (y-2)^2 = 81 and (x-4)^2 + (1-y)^2 = 100. Expanding the squares in these equations we get x^2 - 14 x + 49 + y^2 - 4 y + 4 = 81 and x^2 - 8 x + 16 + y^2 - 2 y + 1 = 100. Collecting terms we have{} x^2 - 14 x + y^2 - 4 y = 28 and x^2 - 8 x + y^2 - 2 y = 83. Subtracting the second equation from the first we get -6x - 2 y = -55, which we solve for y to get y = -3x + 55/2. Substituting this expression into the first equation we get {}x^2 - 14 x + 49 + (-3x + 55/2)^2 - 4 ( -3x + 55/2 ) + 4 = 81, which we expand to get 10·x^2 - 167·x + 701.25 = 83. Solving for x (using the quadratic formula) we get two solutions, x = 11.16 and x = 5.54. Substituting these x values into the second equation we get y^2 - 2·y + 35.2656 = 83, with solution y = 7.98 or y = -5.98; and y^2 - 2·y - 34071/2500 = 83 with solutions y = 10.88 and y = -8.88. This gives us possible solutions (11.16, 7.98), (11.16,-5.98), (5.54, 10.88) and (5.54, -8.88). {}Checking out these solutions with the first equation we see that (11.16, -5.98) and (5.54, 10.88) are in fact solutions, while (11.16, 7.98) and (5.54, 10.88) are not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Write as an equation: The distance from (7, 2) to (x, y) is equal to the distance from (4, 1) to (x, y). Simplify this equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sqrt((x - 7)^2 + (y - 2)^2)) = Sqrt((x - 1)^2 + (y - 1)^2)) (x-7)^2 + (y-2)^2 = (x-4)^2 + (1-y)^2 x^2 - 14 x + 49 + y^2 - 4 y + 4 = x^2 - 8 x + 16 + y^2 - 2 y + 1 -6x - 2y + 36 = 0 y = -3x + 18 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance from (7, 2) to (x, y) is sqrt((x-7)^2 + (y-2)^2), and the distance from (4, 1) to (x, y) is sqrt((x-4)^2 + (1-y)^2). To say that these distances are equal is to say that sqrt((x-7)^2 + (y-2)^2)= sqrt((x-4)^2 + (1-y)^2) Squaring both sides we get (x-7)^2 + (y-2)^2 = (x-4)^2 + (1-y)^2. Expanding the squares we x^2 - 14 x + 49 + y^2 - 4 y + 4 = x^2 - 8 x + 16 + y^2 - 2 y + 1. Subtracting x^2 - 8 x + 16 + y^2 - 2 y + 1 from both sides we get -6x - 2y + 36 = 0. Solving for y we get y = -3x + 18. This is a linear equation, telling us that the set of points (x, y) which are equidistant from (7, 2) and (4, 1) lie along a straight line with slope -3 and y-intercept 18. Recall from basic geometry that the perpendicular bisector of a line segment through two points is the line which is equidistant from those points. Since the slope of the segment from (7,2) to (4,1) is 1/3, we expect that the perpendicular bisector will have slope - 1 / (1/3) = -3, as is the case for the line we have obtained. It is also easy to verify that the line y = -3x + 18 contains the midpoint between (7, 2) and (4, 1): The midpoint is ( (7+4)/2, (2+1)/2 ) = (11/2, 3/2). Substitution will show that this point lies on the line y = -3x + 18. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Write as an equation: The distance from ((7, 4) to (x, y) is equal to the distance from the line y = 2 to (x, y). Simplify this equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sqrt((x - 7)^2 + (y - 4)^2) = y - 2 (x-7)^2 + (y-4)^2 = (y - 2 ) ^ 2 x^2 - 14 x + 49 + y^2 - 8 y + 16 = y^2 - 4 y + 4 x^2 - 14 x + 49 - 4 y + 12 = 0 X^2 - 14x - 4y + 61 = 0 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe distance from (7, 4) to (x, y) is sqrt( (x-7)^2 + (y-4)^2 ). The distance from (x, y) to the line y = 2 lies along the vertical line from (x, y) to y = 2; it is clear from Figure 49 that this distance is | y - 2 |. Thus we have sqrt( (x-7)^2 + (y-4)^2 ) = y - 2. Squaring both sides we get (x-7)^2 + (y-4)^2 = (y - 2 ) ^ 2. Expanding the squares we have x^2 - 14 x + 49 + y^2 - 8 y + 16 = y^2 - 4 y + 4. Subtracting y&2 - 4 y + 4 from both sides we have x^2 - 14 x + 49 - 4 y + 12 = 0. You aren't expected to have known the rest of this solution before, but you need to note the following: Note that this equation is quadratic in x and linear in y. An equation of this form is generally rearranged into the form (y - k) = A * (x - h) ^ 2. In this case we can add 4y - 12 to both sides to get x^2 - 14 x + 49 = 4 ( y - 3). This simplifies to (y - 3) = 1/4 ( x - 7)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Forgot theses steps x^2 - 14 x + 49 = 4 ( y - 3) (y - 3) = 1/4 ( x - 7)^2 ------------------------------------------------ Self-critique Rating:3