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Center (0, 2) a
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Right-most point on curve
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Lower vertex is at (0, 2)
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Illustration for the solution to Problem 5 in Chap 13
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which simplifies to x 2 = 12 This equation solves easily to x = 121/2 or x = 121/2 telling us that the left-most point on the curve is at ( 121/2,2), and the right-most point on the curve is at (121/2,2) The horizontal semi-axis (call it a) is therefore 121/2 units long In the generalized standard form for the equation of an ellipse, the ordered pair (x0,y0) represents the coordinates of the center point We know that this point is (0,2) We ve now figured out these four values: x0 = 0 y0 = 2 a = 121/2 b=4 As you can guess, we chose the names of these parameters so that they d fit neatly into the generalized standard form for the equation of an ellipse That form is (x x0)2/a2 + (y y0)2 /b2 = 1 Plugging in the numbers straightaway, we obtain (x 0)2/(121/2)2 + (y 2)2 /42 = 1 which simplifies to x 2/12 + (y 2)2/16 = 1 At last, we ve found the standard-form equation for the ellipse graphed in Fig 13-12! If you re skeptical and ambitious, you ll demand proof that this equation is equivalent to the one we got in the solution to Problem 3 Why not demonstrate that fact as an extra-credit exercise Start with x 2/12 + (y 2)2/16 = 1 and morph it into x 2 + y 2 = (6 + y)2 /4
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532 Worked-Out Solutions to Exercises: 11-19
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6 Here s the original equation again, for reference: x 2 + 9y 2 = 9 We can divide this equation through by 9 to obtain x 2 /9 + y 2 = 1 This equation is in the standard form for an ellipse We recall that the general version is (x x0)2 /a2 + (y y0)2 /b2 = 1 where x0 and y0 are the coordinates of the center, a is the length of the horizontal semiaxis, and b is the length of the vertical semi-axis In this case, we have x0 = 0 y0 = 0 a = 91/2 = 3 b = 11/2 = 1 The center is at (0,0) The horizontal semi-axis measures 3 units The vertical semi-axis measures 1 unit We can now sketch the graph as shown in Fig B-9
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Each axis increment is 1/2 unit b=1 b a x a=3
(x 0 , y 0 ) = (0, 0)
Figure B-9
Illustration for the solution to Problem 6 in Chap 13 Each axis division represents 1/2 unit
13
7 Here s the original equation again, for reference: x 2 + y 2 + 2x 2y + 2 = 4 It takes some mathematical intuition to see how this can be morphed into the standard form for a conic section We can split it into the sum of two trinomials as (x 2 + 2x + 1) + (y 2 2y + 1) = 4 The parentheses, while not technically necessary, clarify the identities of the trinomials Both of these trinomials happen to be perfect squares They can be factored to get (x + 1)2 + (y 1)2 = 4 which is in the standard form for a circle We remember that the general equation for a circle is (x x0)2 + (y y0)2 = r2 where (x0,y0) are the coordinates of the center, and r the radius In this particular example, we have x0 = 1 y0 = 1 r = 41/2 = 2 The center is at ( 1,1) The radius is 2 units We can now sketch the graph as shown in Fig B-10
Figure B-10
Illustration for the solution to Problem 7 in Chap 13 Each axis division represents 1/2 unit
r r=2
(x0, y0) = ( 1, 1)
Each axis increment is 1/2 unit
534 Worked-Out Solutions to Exercises: 11-19
8 Here s the original equation again, for reference: x 2 y 2 + 2x + 2y = 4 As in Problem 7, we must rely on our algebra experience to see how this can be transformed into an equation that s in the standard form for a conic section We can split it into a difference between two trinomials as (x 2 + 2x + 1) (y 2 2y + 1) = 4 In this equation, the parentheses are necessary! The trinomials can be factored exactly as they were in the solution to Problem 7; when we do that, we obtain (x + 1)2 (y 1)2 = 4 We can divide through by 4, getting (x + 1)2 /4 (y 1)2 /4 = 1 This equation is in the standard form for a hyperbola The general version, as we ve learned, is (x x0)2 /a2 (y y0)2 /b2 = 1 where (x0,y0) are the coordinates of the center, a is the length of the horizontal semi-axis, and b is the length of the vertical semi-axis In this case, we have x0 = 1 y0 = 1 a = 41/2 = 2 b = 41/2 = 2 The center is at ( 1,1) The horizontal and vertical semi-axes are both 2 units long With this information, we can sketch the graph as shown in Fig B-11 9 Here s the original equation again, for reference: x 2 3x y + 3 = 1 We can subtract 1 from each side to get x 2 3x y + 2 = 0 Adding y to each side and then transposing the sides left-to-right yields y = x 2 3x + 2
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