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The Projective Plane 8.5 Figures 8.13 and 8. 14 were made by taking Alberti's veil to be the (x,z)-plane in (x, y, z)-space, with the "eye" at (0, -4,4) viewing the (x, y)-plane. 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x,y,0), and 141 hence show that this line meets the veil where 4x' 4y Z = y' +4 y' 4 843 Renaming the coordinates x, z in the veil as X, Y respectively, show that 4X 4Y V' 4 Y 4-Y 8.4.4 Deduce from Exercise 8.4.3 that the points (x', y') on the parabola y = have image on the veil (Y-2) = 1, X2 + 4 and check that this is the ellipse shown in Figure 8.13. 8.5 The Projective Plane The way in which projective geometry allows infinity to be put on the same footing of the horizon in a picture, which is a line like any other. But what, math ematically speaking, is this line we see? We can model the situation math as the finite points of the plane is intuitively clear when one thinks ha the plane z = -1 in th - eve 8.4.2 Find the parametric equations of the line from (0, -4,4) to (x', y', 0) and hence show that this line meets the veil where: 4x 4y z=_ y4 y'4 х — х' у-у' Z 0 0 x'0 y'4 4 х — х' у —у' Z .. (1) = х' y' 4 -4 х — х' у-у' х' y'4 x'(y-' x(y' 4) x'(4 — х(у' + 4) — х'(у' + 4+у-у) %3D0 — х(у' + 4) %3 х"(4 + у) x'(4y) (2) y' 4) у-у' z y'4 -4 z(y'+ 4)4(y-y') 4y'y - (3) Z= y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4x 4y' ;2= y' 4 y'4 0 у-у' y'4 Suppose = k -4 yy'ky'4) y k(y' +4) y z=-4k Also suppose y+4 4(1 a) x ax', z ay',4 = ay' + 4a »y' ; z = 4(1 -a) a (k 1)x'; y (k+1)y' 4k;z = -4k Hence parametric equation is given by: x =

The Projective Plane 8.5 Figures 8.13 and 8. 14 were made by taking Alberti's veil to be the (x,z)-plane in (x, y, z)-space, with the "eye" at (0, -4,4) viewing the (x, y)-plane. 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x,y,0), and 141 hence show that this line meets the veil where 4x' 4y Z = y' +4 y' 4 843 Renaming the coordinates x, z in the veil as X, Y respectively, show that 4X 4Y V' 4 Y 4-Y 8.4.4 Deduce from Exercise 8.4.3 that the points (x', y') on the parabola y = have image on the veil (Y-2) = 1, X2 + 4 and check that this is the ellipse shown in Figure 8.13. 8.5 The Projective Plane The way in which projective geometry allows infinity to be put on the same footing of the horizon in a picture, which is a line like any other. But what, math ematically speaking, is this line we see? We can model the situation math as the finite points of the plane is intuitively clear when one thinks ha the plane z = -1 in th - eve 8.4.2 Find the parametric equations of the line from (0, -4,4) to (x', y', 0) and hence show that this line meets the veil where: 4x 4y z=_ y4 y'4 х — х' у-у' Z 0 0 x'0 y'4 4 х — х' у —у' Z .. (1) = х' y' 4 -4 х — х' у-у' х' y'4 x'(y-' x(y' 4) x'(4 — х(у' + 4) — х'(у' + 4+у-у) %3D0 — х(у' + 4) %3 х"(4 + у) x'(4y) (2) y' 4) у-у' z y'4 -4 z(y'+ 4)4(y-y') 4y'y - (3) Z= y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4x 4y' ;2= y' 4 y'4 0 у-у' y'4 Suppose = k -4 yy'ky'4) y k(y' +4) y z=-4k Also suppose y+4 4(1 a) x ax', z ay',4 = ay' + 4a »y' ; z = 4(1 -a) a (k 1)x'; y (k+1)y' 4k;z = -4k Hence parametric equation is given by: x =

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I need ass istance with this for History of math 8.4.3

 

I have attached how I did 8.4.2 for reference

The Projective Plane 8.5 Figures 8.13 and 8. 14 were made by taking Alberti's veil to be the (x,z)-plane in (x, y, z)-space, with the "eye" at (0, -4,4) viewing the (x, y)-plane. 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x,y,0), and 141 hence show that this line meets the veil where 4x' 4y Z = y' +4 y' 4 843 Renaming the coordinates x, z in the veil as X, Y respectively, show that 4X 4Y V' 4 Y 4-Y 8.4.4 Deduce from Exercise 8.4.3 that the points (x', y') on the parabola y = have image on the veil (Y-2) = 1, X2 + 4 and check that this is the ellipse shown in Figure 8.13. 8.5 The Projective Plane The way in which projective geometry allows infinity to be put on the same footing of the horizon in a picture, which is a line like any other. But what, math ematically speaking, is this line we see? We can model the situation math as the finite points of the plane is intuitively clear when one thinks ha the plane z = -1 in th - eve
Transcribed Image Text:The Projective Plane 8.5 Figures 8.13 and 8. 14 were made by taking Alberti's veil to be the (x,z)-plane in (x, y, z)-space, with the "eye" at (0, -4,4) viewing the (x, y)-plane. 8.4.2 Find the parametric equations of the line from (0,-4,4) to (x,y,0), and 141 hence show that this line meets the veil where 4x' 4y Z = y' +4 y' 4 843 Renaming the coordinates x, z in the veil as X, Y respectively, show that 4X 4Y V' 4 Y 4-Y 8.4.4 Deduce from Exercise 8.4.3 that the points (x', y') on the parabola y = have image on the veil (Y-2) = 1, X2 + 4 and check that this is the ellipse shown in Figure 8.13. 8.5 The Projective Plane The way in which projective geometry allows infinity to be put on the same footing of the horizon in a picture, which is a line like any other. But what, math ematically speaking, is this line we see? We can model the situation math as the finite points of the plane is intuitively clear when one thinks ha the plane z = -1 in th - eve
8.4.2 Find the parametric equations of the line from (0, -4,4) to (x', y', 0) and hence show that this line meets the veil where: 4x 4y z=_ y4 y'4 х — х' у-у' Z 0 0 x'0 y'4 4 х — х' у —у' Z .. (1) = х' y' 4 -4 х — х' у-у' х' y'4 x'(y-' x(y' 4) x'(4 — х(у' + 4) — х'(у' + 4+у-у) %3D0 — х(у' + 4) %3 х"(4 + у) x'(4y) (2) y' 4) у-у' z y'4 -4 z(y'+ 4)4(y-y') 4y'y - (3) Z= y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4x 4y' ;2= y' 4 y'4 0 у-у' y'4 Suppose = k -4 yy'ky'4) y k(y' +4) y z=-4k Also suppose y+4 4(1 a) x ax', z ay',4 = ay' + 4a »y' ; z = 4(1 -a) a (k 1)x'; y (k+1)y' 4k;z = -4k Hence parametric equation is given by: x =
Transcribed Image Text:8.4.2 Find the parametric equations of the line from (0, -4,4) to (x', y', 0) and hence show that this line meets the veil where: 4x 4y z=_ y4 y'4 х — х' у-у' Z 0 0 x'0 y'4 4 х — х' у —у' Z .. (1) = х' y' 4 -4 х — х' у-у' х' y'4 x'(y-' x(y' 4) x'(4 — х(у' + 4) — х'(у' + 4+у-у) %3D0 — х(у' + 4) %3 х"(4 + у) x'(4y) (2) y' 4) у-у' z y'4 -4 z(y'+ 4)4(y-y') 4y'y - (3) Z= y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4x 4y' ;2= y' 4 y'4 0 у-у' y'4 Suppose = k -4 yy'ky'4) y k(y' +4) y z=-4k Also suppose y+4 4(1 a) x ax', z ay',4 = ay' + 4a »y' ; z = 4(1 -a) a (k 1)x'; y (k+1)y' 4k;z = -4k Hence parametric equation is given by: x =
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