 Home
 Products
 Integration
 Tutorial
 Barcode FAQ
 Purchase
 Company
qr code vb.net open source Point reflector line y 6 4 Inverse relation x 6 4 4 6 in .NET framework
12 USS Code 39 Scanner In VS .NET Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in VS .NET applications. Draw Code 39 In Visual Studio .NET Using Barcode maker for .NET Control to generate, create ANSI/AIM Code 39 image in Visual Studio .NET applications. Point reflector line y 6 4 Inverse relation x 6 4 4 6
Recognizing Code 39 In .NET Framework Using Barcode decoder for .NET framework Control to read, scan read, scan image in .NET framework applications. Generating Bar Code In VS .NET Using Barcode generation for .NET Control to generate, create barcode image in .NET framework applications. 4 6 Scan Barcode In Visual Studio .NET Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications. Generating Code 39 Full ASCII In C#.NET Using Barcode creation for .NET Control to generate, create ANSI/AIM Code 39 image in .NET applications. Original relation
Code 39 Extended Generation In Visual Studio .NET Using Barcode printer for ASP.NET Control to generate, create Code 3/9 image in ASP.NET applications. Code 39 Extended Maker In Visual Basic .NET Using Barcode drawer for .NET Control to generate, create ANSI/AIM Code 39 image in Visual Studio .NET applications. Figure B3 Paint Linear 1D Barcode In .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Linear image in .NET applications. Bar Code Creator In Visual Studio .NET Using Barcode encoder for VS .NET Control to generate, create barcode image in Visual Studio .NET applications. Illustration for the solution to Problem 7 in Chap 12
Draw UPC Code In .NET Framework Using Barcode drawer for VS .NET Control to generate, create UPC A image in .NET framework applications. British Royal Mail 4State Customer Code Generation In .NET Using Barcode maker for Visual Studio .NET Control to generate, create Royal Mail Barcode image in .NET applications. Inverse of original restricted relation
Code 39 Extended Generation In .NET Using Barcode encoder for Reporting Service Control to generate, create Code 39 image in Reporting Service applications. Barcode Creator In Java Using Barcode creation for Java Control to generate, create barcode image in Java applications. y 6 4 Generate Barcode In VB.NET Using Barcode generation for VS .NET Control to generate, create barcode image in .NET framework applications. Bar Code Creation In None Using Barcode generation for Office Excel Control to generate, create barcode image in Microsoft Excel applications. x 6 4 2 2 Movable 4 vertical line 6 Original restricted relation 4 6
Bar Code Creation In None Using Barcode drawer for Font Control to generate, create barcode image in Font applications. ANSI/AIM Code 39 Printer In None Using Barcode maker for Software Control to generate, create Code39 image in Software applications. Figure B4 EAN13 Printer In .NET Using Barcode maker for Reporting Service Control to generate, create EAN / UCC  13 image in Reporting Service applications. Barcode Printer In C# Using Barcode drawer for Visual Studio .NET Control to generate, create barcode image in .NET applications. Illustration for the solution to Problem 8 in Chap 12
526 WorkedOut Solutions to Exercises: 1119 y 6 4 2 x 6 4 2 4 6 Movable vertical line
Original restricted relation
4 6 Inverse of original restricted relation
Figure B5 Illustration for the solution to Problem 9 in Chap 12
9 Figure B5 shows the graph of the original relation with its domain restricted to the reals smaller than or equal to 2 ( gray curve) and its inverse (black curve) The curve for the inverse relation is a halfhyperbola that opens downward and intersects the y axis at (0, 2) A movable vertical line never intersects the black curve at more than one point Therefore, the inverse relation f 1 (x) = (4x 2 /9 + 4)1/2 is a true function of x 10 Figure B6 shows the graph of the original relation with its range restricted to the nonnegative reals (pair of gray curves) and the graph of its inverse (pair of black curves) A movable vertical line never intersects the graph of the original restricted relation at more than one point, so it s a true function Whenever the movable vertical line intersects the graph of the inverse, that line crosses both curves On this basis, we know that the inverse of the original restricted relation is not a true function 13
Original restricted relation y 6 4 2
2 2 4 6 Inverse of x original restricted relation
Point reflector line
Movable vertical line
Figure B6 Illustration for the solution to Problem 10 in Chap 12
13
1 If the plane passes through the point where the apexes of the two cones meet, the intersection is a point, a straight line, or a pair of lines that intersect Here s how the situations break down: If the plane is slanted so that we d get a circle or ellipse if the plane didn t pass through the apexes, then we get a single point ( figs B7A and B) If the plane slants so that we d get a parabola if the plane didn t pass through the apexes, then we get a straight line ( fig B7C) If the plane slants so that we d get a hyperbola if the plane didn t pass through the apexes, we get a pair of lines that cross ( fig B7D) 2 When we examine the graph of the ellipse, we see that the lower focus is at (0,0) and the lower vertex is at (0, 2) Therefore, the distance between the lower focus and the lower vertex is 2 units, telling us that the focal length is f =2 528 WorkedOut Solutions to Exercises: 1119 Double circular cone Double circular cone
Flat plane
Flat plane
Flat plane
Flat plane
Double circular cone
Double circular cone
Figure B7 Illustrations for the solution to Problem 1 in Chap 13
We can also see that the distance between the lower vertex and the directrix is 4 units, because the lower vertex is at (0, 2) and the directrix passes through (0, 6) Therefore f /e = 4 Combining the above two equations and solving, we get e = 1/2 That s the eccentricity of the ellipse 3 When we solved Problem 2, we determined that f = 2 and e = 1/2 Plugging these values into the formulas relating the parameters of the ellipse, we obtain u = 2 + 2 /(1/2) + y and x 2 + y 2 = [(1/2) u]2 13
These simplify to u=6+y and x 2 + y 2 = u2/4 Let s substitute the quantity (6 + y) for u in the second equation above That gives us x 2 + y 2 = (6 + y)2/4 This is an equation for our ellipse, although it s not in standard form 4 We ve been told that both foci and both vertices lie on the y axis, so x = 0 at the lower vertex and also at the upper vertex If we plug x = 0 into the equation we got in the solution to Problem 3, we ll be left with an equation that tells us the y values at both vertices of the ellipse Here it is y 2 = (6 + y)2/4 Multiplying through by 4, we obtain 4y 2 = (6 + y)2 When we multiply out the squared binomial on the righthand side, we get 4y 2 = 36 + 12y + y 2 This is a quadratic equation Let s morph it with algebra into the standard form for a quadratic That gives us 3y 2 12y 36 = 0 which factors into (3y + 6)(y 6) = 0 The solutions are therefore y = 2 or y=6 The first solution corresponds to the point (0, 2) on the ellipse The second solution corresponds to (0,6) on the ellipse We already know that the coordinates of the lower vertex are (0, 2), so the coordinates of the upper vertex must be (0,6) 530 WorkedOut Solutions to Exercises: 1119 Now let s find the coordinates of the upper focus We already know that the lower focus is at (0,0) We also know that the lower vertex is at (0, 2), which is 2 units below the lower focus on the y axis Because all ellipses are symmetrical with respect to the foci, the upper focus must be 2 units below the upper vertex on the y axis Therefore, the coordinates of the upper focus must be (0,4) 5 We can find the coordinates of our ellipse s center by averaging the coordinates of the foci The lower focus is at (0,0), while the upper focus is at (0,4) Therefore, the center must be at (0,2) The length of the vertical semiaxis (let s call it b) is the distance between the center and the lower vertex, as shown in Fig B8 The center is at (0,2) and the lower vertex is at (0, 2), so the vertical semiaxis is 4 units long (We get the same result if we use the center and the upper vertex) To find the length of the horizontal semiaxis, we must know the coordinates of either the leftmost point or the rightmost point on the ellipse These are the points where the horizontal line y = 2, which passes through the center of the ellipse, intersects the curve To find these points, we can plug y = 2 into the equation we got for the ellipse when we solved Problem 3 When we make the substitutions, we obtain x 2 + 22 = (6 + 2)2/4 Upper vertex is at (0, 6)

