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which simplifies to (x + 2)2 /a2 9/b2 + (z 4)2 /c2 = 0 If we add the quantity 9/b2 to both sides, we obtain (x + 2)2 /a2 + (z 4)2 /c2 = 9/b2 Now we can divide through by the quantity 9/b2, getting (x + 2)2 /(9a2 /b2) + (z 4)2 /(9c2 /b2) = 1 which can also be expressed as (x + 2)2 /(3a/b)2 + (z 4)2 /(3c/b)2 = 1 This is a generalized equation for an ellipse in the Cartesian plane, where the variables are x and z The center of the ellipse has the coordinates (x0,z0) = ( 2,4) The semi-axes have lengths 3a/b and 3c/b 10 Here, once again, is the generalized equation that we derived for the elliptic cone described in Problem 8: (x + 2)2 /a2 (y 3)2 /b2 + (z 4)2 /c2 = 0 This cone intersects the xy plane in a curve where z = 0 at every point If we set z = 0 in the above equation, we get (x + 2)2 /a2 (y 3)2 /b2 + (0 4)2 /c2 = 0 We can simplify this equation to (x + 2)2 /a2 (y 3)2 /b2 + 16/c2 = 0 If we subtract the quantity 16/c2 from both sides, we obtain (x + 2)2 /a2 (y 3)2 /b2 = 16/c2 Dividing through by the quantity 16/c2, we obtain (x + 2)2 /(16a2 /c2) (y 3)2 /(16b2 /c2) = 1 which can be rewritten as (x + 2)2 /(4a/c)2 (y 3)2 /(4b/c)2 = 1
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Finally, we can multiply the entire equation through by 1 to obtain (y 3)2 /(4b/c)2 (x + 2)2 /(4a/c)2 = 1 This is a generalized equation for a hyperbola in the Cartesian xy plane It s oriented differently than the hyperbolas in Chap 13, however Instead of opening in the positive and negative x directions as the hyperbolas in Chap 13 do, this pair of curves opens in the positive and negative y directions The coordinates of the center are (x0,y0) = ( 2,3) The lengths of the semi-axes are 4a/c and 4b/c
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1 We have a three-way equation for a straight line We want to find the preferred (lowest) direction numbers, and then find the direction vector based on those numbers We also want to know the coordinates of a specific point on the line; any point will suffice Stated again for reference, our three-way equation is x 1 = y 2 = z 4 This equation is in the standard symmetric form for a straight line in Cartesian xyz space, so we don t have to manipulate anything The generalized standard-form symmetric equation is (x x0)/a = (y y0)/b = (z z0)/c where (x0,y0,z0) are the coordinates of a specific point on the line, and a, b, and c are the direction numbers Comparing the symmetric equation we ve been given with the generalized form, we can see that x0 = 1 y0 = 2 z0 = 4 This tells us that (1,2,4) is a point on the line We can also see that a=1 b=1 c=1 so the direction numbers are (1,1,1) This set of numbers is in lowest form, because there s no common divisor other than 1 We can write down a standard-form direction vector m from these values as m=i+j+k
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2 Stated again for reference, the three-way equation for our line is 4x = 5y = 6z Dividing the entire equation through by 60, we obtain x /15 = y /12 = z/10 The generalized standard-form symmetric equation is (x x0)/a = (y y0)/b = (z z0)/c where (x0,y0,z0) are the coordinates of a specific point on the line, and a, b, and c are the direction numbers In this case, we have x0 = 0 y0 = 0 z0 = 0 This tells us that the origin (0,0,0) is on the line We also have a = 15 b = 12 c = 10 so the line s direction numbers are (15,12,10) This ordered triple is in lowest form, so the direction vector m is m = 15i + 12j + 10k 3 We ve been given the symmetric equation (x 2)/3 = (4y 8)/4 = (z + 5)/( 2) We can divide out the middle portion to get (x 2)/3 = y 2 = (z + 5)/( 2) This equation is in the standard symmetric form Once again, the generalized standardform symmetric equation is (x x0)/a = (y y0)/b = (z z0)/c In this situation, we have x0 = 2 y0 = 2 z0 = 5
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so we know that the point (2,2, 5) is on the line We can also see that a=3 b=1 c = 2 so the line s direction numbers are (3,1, 2) This ordered triple is in lowest form (We could divide it through by 1, but then we d get two negative elements instead of only one) The direction vector m is therefore m = 3i + j 2k 4 Stated again for convenience, the parametric equations for our object are x = 4 y=t z = t 2 1 The first equation tells us that the object lies in the plane x = 4, which is perpendicular to the x axis, parallel to the yz plane, and 4 units distant from the yz plane on the x side We can draw projections of the coordinate y and z axes into the plane x = 4, obtaining a Cartesian yz grid In that system, our object is a parabola defined by y=t and z = t2 1 Substituting y for t in the second equation gives us z = y 2 1 Figure B-32 is a graph of this curve as it looks when we see the plane x = 4 broadside from a point on the positive x axis at a considerable distance from the origin Figure B-32
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