x2have the 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y image on the veil (Y 2)2 X2 1 4 And check that this is the ellipse shown in Figure 8.13 x 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 у 4 Y that this line meets the veil where: 4- Y' 4x' 4y' Solve for y as follows: X = Z = y'4 y4 4y Y(y' 4) 4y Y = y' 4 х — х' у — у' z-0 х' — 0 y' 4 0 4 4y' 4Y = 4y' - y'Y y'Y4Y х — х' У — у" Z 4Y y'(4- Y) (1) y' 4 у — у" х' -4 4Y y= 4 Y х — х' х' y' 4 Solve for x as follows: — х(у' + 4) — х'(у' + 4) — х'(у — у') 4c 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) — 0 — х(у' + 4) — х'(4 + у) 4x' 4Y = 4x + - 4Y x'(4y (y' + 4) +4 (2) X = 4XY 4X = 4x'> 4XY + 4X(4 — Ү) = 4x У — у" y' 4 Z 4 Y 4 Y -4 XҮ + X(4 — Ү) = x' 4 Y XҮ + 4X — XҮ = x — 2(y' + 4) — -4(у — у') 4 Y 4(y' — у) 4X x' = Z 4 Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4X 4Y Hence it is proved that x' = y 4-ү' 4x — х %— 4y' ; У 4-Yг %3D 0 : Z= y'4 y'4 х-х' Suppose у-у' y'+4 = k -4 x' х 3 (k+ 1)x' y - y' ky 4) y k(y' + 4) + y' z -4k Also suppose у' у'+4 х1 4(1 a) х — ах',z 3 аy',4 — аy' + 4а — у' — ; z = 4(1 - a) a Hence parametric equation is given by: x (k 1)x'; y = (k + 1)y'+ 4k; z = -4k
x2have the 8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y image on the veil (Y 2)2 X2 1 4 And check that this is the ellipse shown in Figure 8.13 x 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 у 4 Y that this line meets the veil where: 4- Y' 4x' 4y' Solve for y as follows: X = Z = y'4 y4 4y Y(y' 4) 4y Y = y' 4 х — х' у — у' z-0 х' — 0 y' 4 0 4 4y' 4Y = 4y' - y'Y y'Y4Y х — х' У — у" Z 4Y y'(4- Y) (1) y' 4 у — у" х' -4 4Y y= 4 Y х — х' х' y' 4 Solve for x as follows: — х(у' + 4) — х'(у' + 4) — х'(у — у') 4c 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) — 0 — х(у' + 4) — х'(4 + у) 4x' 4Y = 4x + - 4Y x'(4y (y' + 4) +4 (2) X = 4XY 4X = 4x'> 4XY + 4X(4 — Ү) = 4x У — у" y' 4 Z 4 Y 4 Y -4 XҮ + X(4 — Ү) = x' 4 Y XҮ + 4X — XҮ = x — 2(y' + 4) — -4(у — у') 4 Y 4(y' — у) 4X x' = Z 4 Y y'4 The line (1) meets the veil at y = 0 hence (2) (3) 4X 4Y Hence it is proved that x' = y 4-ү' 4x — х %— 4y' ; У 4-Yг %3D 0 : Z= y'4 y'4 х-х' Suppose у-у' y'+4 = k -4 x' х 3 (k+ 1)x' y - y' ky 4) y k(y' + 4) + y' z -4k Also suppose у' у'+4 х1 4(1 a) х — ах',z 3 аy',4 — аy' + 4а — у' — ; z = 4(1 - a) a Hence parametric equation is given by: x (k 1)x'; y = (k + 1)y'+ 4k; z = -4k
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.3: Hyperbolas
Problem 43E
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Question
need help w/8.4.4
I have attached 8.4.3 for reference
Transcribed Image Text:x2have the
8.4.4 Deduce from exercise 8.4.3 that the points (x', y') on the parabola y
image on the veil
(Y 2)2
X2
1
4
And check that this is the ellipse shown in Figure
8.13
x
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
у
4 Y
that this line meets the veil where:
4- Y'
4x'
4y'
Solve for y as follows:
X =
Z =
y'4
y4
4y
Y(y' 4) 4y
Y =
y' 4
х — х'
у — у'
z-0
х' — 0
y' 4
0 4
4y' 4Y = 4y' - y'Y
y'Y4Y
х — х'
У — у"
Z
4Y y'(4- Y)
(1)
y' 4
у — у"
х'
-4
4Y
y=
4 Y
х — х'
х'
y' 4
Solve for x as follows:
— х(у' + 4) — х'(у' + 4) — х'(у — у')
4c
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) — 0
— х(у' + 4) — х'(4 + у)
4x'
4Y
= 4x
+
-
4Y
x'(4y
(y' + 4)
+4
(2)
X =
4XY
4X = 4x'>
4XY + 4X(4 — Ү)
= 4x
У — у"
y' 4
Z
4 Y
4 Y
-4
XҮ + X(4 — Ү)
= x'
4 Y
XҮ + 4X — XҮ
= x
— 2(y' + 4) — -4(у — у')
4 Y
4(y' — у)
4X
x' =
Z
4 Y
y'4
The line (1) meets the veil at y = 0 hence (2) (3)
4X
4Y
Hence it is proved that x' =
y
4-ү'
4x
— х %—
4y'
; У
4-Yг
%3D 0
: Z=
y'4
y'4
х-х'
Suppose
у-у'
y'+4
= k
-4
x'
х 3 (k+ 1)x'
y - y' ky
4) y k(y' + 4) + y'
z -4k
Also suppose
у'
у'+4
х1
4(1 a)
х — ах',z 3 аy',4 — аy' + 4а — у' —
; z = 4(1 - a)
a
Hence parametric equation is given by: x (k 1)x'; y = (k + 1)y'+ 4k; z = -4k
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