Find the shortest distance from the origin to the hyperbola x2 8xy 7y2 225, z 0. in .NET

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8.26. Find the shortest distance from the origin to the hyperbola x2 8xy 7y2 225, z 0.
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We must nd the minimum value of x2 y2 (the square of the distance from the origin to any point in the xy plane) subject to the constraint x2 8xy 7y2 225. According to the method of Lagrange multipliers, we consider  x2 8xy 7y2 225  x2 y2 . Then x 2x 8y 2x 0 y 8x 14y 2y 0 or or 1  1 x 4y 0 2 4x  7 y 0
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From (1) and (2), since x; y 6 0; 0 , we must have    1 4    0; i:e:; 2 8 9 0 or  4  7
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Case 1:  1. From (1) or (2), x 2y and substitution in x2 8xy 7y2 225 yields 5y2 225, for which no real solution exists. Case 2:  9. From (1) or (2), y 2x and substitution in x2 8xy 7y2 225 yields 45x2 225. p Then x2 5; y2 4x2 20 and so x2 y2 25. Thus the required shortest distance is 25 5.
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(a) Find the maximum and minimum values of x2 y2 z2 subject to the constraint conditions x2 =4 y2 =5 z2 =25 1 and z x y. (b) Give a geometric interpretation of the result in (a).
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x2 y2 (a) We must nd the extrema of F x2 y2 z2 subject to the constraint conditions 1 2 4 5 z 1 0 and 2 x y z 0. In this case we use two Lagrange multipliers 1 ; 2 and consider 25 the function ! x2 y2 z2 2 2 2 1 2 x y z G F 1 1 2 2 x y z 1 4 5 25 Taking the partial derivatives of G with respect to x; y; z and setting them equal to zero, we nd Gx 2x 1 x 2 0; 2 Gy 2y 21 y 2 0; 5 Gx 2z 21 z 2 0 25 1
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Solving these equations for x; y; z, we nd x 22 ; 1 4 y 52 ; 21 10 z 252 21 50 2
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From the second constraint condition, x y z 0, we obtain on division by 2 , assumed different from zero (this is justi ed since otherwise we would have x 0; y 0; z 0, which would not satisfy the rst constraint condition), the result 2 5 25 0 1 4 21 10 21 50 Multiplying both sides by 2 1 4 1 5 1 25 and simplifying yields 172 2451 750 0 1 from which 1 10 or 75=17. Case 1: 1 10. From (2), x 1 2 ; y 1 2 ; z 5 2 . pSubstituting in the rst constraint condition, x2 =4 y2 =5 3 2 6 2 z =25 1, yields 2 180=19 or 2 6 5=19. This gives the two critical points 2 p p p p p p 2 5=19; 3 5=19; 5 5=19 ; 2 5=19; 3 5=19; 5 5=19 The value of x2 y2 z2 corresponding to these critical points is 20 45 125 =19 10. Case 2: 1 75=17: From (2), x 34 2 ; y 17 2 ; z 17 2p Substituting in the rst constraint condition, . 7 4 28 2 x =4 y2 =5 z2 =25 1, yields 2 140= 17 646 which gives the critical points p p p p p p 40= 646; 35 646; 5= 646 ; 40= 646; 35= 646; 5= 646 The value of x2 y2 z2 corresponding to these is 1600 1225 25 =646 75=17. Thus, the required maximum value is 10 and the minimum value is 75/17. (b) Since x2 y2 z2 represents the square of the distance of x; y; z from the origin 0; 0; 0 , the problem is equivalent to determining the largest and smallest distances from the origin to the curve of intersec2 tion of the ellipsoid x2 =4 yp z2 =25 1 and the plane z x y. Since this curve is an ellipse, we =5 p have the interpretation that 10 and 75=17 are the lengths of the semi-major and semi-minor axes of this ellipse. The fact that the maximum and minimum values happen to be given by 1 in both Case 1 and Case 2 is more than a coincidence. It follows, in fact, on multiplying equations (1) by x, y, and z in succession and adding, for we then obtain 2x2 1 x2 2 y2 2 z2 2 x 2y2 1 2 y 2z2 1 2 z 0 2 5 25
2 2 2
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