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qr code vb.net open source Leftmost point on curve Line y = 2 in VS .NET
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Data Matrix 2d Barcode Printer In None Using Barcode maker for Software Control to generate, create Data Matrix 2d barcode image in Software applications. Scanning EAN13 In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. which simplifies to x 2 = 12 This equation solves easily to x = 121/2 or x = 121/2 telling us that the leftmost point on the curve is at ( 121/2,2), and the rightmost point on the curve is at (121/2,2) The horizontal semiaxis (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 standardform equation for the ellipse graphed in Fig 1312! 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 extracredit exercise Start with x 2/12 + (y 2)2/16 = 1 and morph it into x 2 + y 2 = (6 + y)2 /4 Code 128 Code Set A Generator In Java Using Barcode encoder for BIRT reports Control to generate, create Code128 image in BIRT applications. EAN 128 Maker In Java Using Barcode creation for BIRT Control to generate, create GTIN  128 image in BIRT applications. 532 WorkedOut Solutions to Exercises: 1119 Printing Barcode In None Using Barcode generation for Font Control to generate, create bar code image in Font applications. EAN13 Supplement 5 Recognizer In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. 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 semiaxis 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 semiaxis measures 3 units The vertical semiaxis measures 1 unit We can now sketch the graph as shown in Fig B9 Universal Product Code Version A Encoder In Java Using Barcode creation for Java Control to generate, create UPCA Supplement 5 image in Java applications. ANSI/AIM Code 128 Encoder In VS .NET Using Barcode generator for ASP.NET Control to generate, create Code 128 image in ASP.NET applications. Each axis increment is 1/2 unit b=1 b a x a=3
(x 0 , y 0 ) = (0, 0) Figure B9 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 B10 Figure B10 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 WorkedOut Solutions to Exercises: 1119 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 semiaxis, and b is the length of the vertical semiaxis 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 semiaxes are both 2 units long With this information, we can sketch the graph as shown in Fig B11 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 lefttoright yields y = x 2 3x + 2

