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S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola

S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.4: Plane Curves And Parametric Equations
Problem 32E
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Can you help me with 8.4.4, I dont know what its asking or how to solve

 

S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola
Transcribed Image Text:S-8.4.2-8.4.4,8.7.1,8.7.2,9.2.2,9.2.5,9.5.1,9.5.2 8.4.3 Renaming the coordinates x, z in the veil as X, Y respectively show that 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x', y',0) and hence show 4X 4Y y= 4 Y 4 Y that this line meets the veil where: 4y 4x' х%3 y'4 Solve for y as follows: Z= y 4 4y' — Yу' + 4) — 4y' Y = у — у' y'4 х — х' z- 0 x'- 0 0 4 y'4 y'Y4Y 4y' » 4Y = 4y' - y'Y х — х' У — у' Z 4Y y'(4 Y) (1) y'4 У — у" 4Y х — х' y= 4 Y' y'4 Solve for x as follows: — х(у + 4) — х'(у' + 4) — х'(у — у') 4x and obtain the equation for x' below: y'4 Substitute the equation obtained for y' in x = — х(у' + 4) — х'(у'+4+у—у) %3D 0 — х(у' + 4) — х'(4 + у) 4x' 4Y X 4Y x'(4y) +4 2) X (y' 4) 4XY 4X(4 Y) 4XY +4X 4 - Y 4x' = 4x' У — у" y'4 Z 4 Y -4 XҮ + 4X — ХY = x XҮ + X(4 — Ү) = x' > — 2(y' + 4) — -4(у — у') 4 Y 4 Y 4(y' — у) 4X x' = (3) Z= 4 - Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4Y Hence it is proved that x' = y = 4-Y 4y y'4 4-Y 4x' — х — ;Z= ;y = 0 y'4 x-x' Suppose у-у' у'+4 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y = x-have the image on the veil = k x = (k + 1)x' (Ү— 2)2 X2 у — у' %3D k(y' + 4) — у %3D k(y' + 4) + у' 4 And check that this is the ellipse shown in Figure z = -4k 8.13 Also suppose = a y'+4 y' 4(1 — а) х — ах',z — ау',4 — ау' + 4a — у' — ;z = 4(1 - a) а Hence parametric equation is given by: x (k + 1)x'; y = (k + 1)y' 4k; z = -4k Figure 8.13: The parabola
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