Transformation of Coordinates PDF

# Transformation of Coordinates ## Chapter 2: Transformation of Coordinates **8. The coordinates of a point depend upon the origin and the axes of coordinates chosen.** If a point be given in position, then its coordinates referred to one set of axes will change as soon as a new set of axes be tak...

# Transformation of Coordinates ## Chapter 2: Transformation of Coordinates **8. The coordinates of a point depend upon the origin and the axes of coordinates chosen.** If a point be given in position, then its coordinates referred to one set of axes will change as soon as a new set of axes be taken. It is evident that the equation of a curve will also change by such a transformation. But if we know the equation of the curve referred to one set of axes, we can deduce the equation referred to another set of axes. When we pass from one set of axes to another, the process is known as the Transformation of Coordinates. **9. To change the origin of coordinates without changing the direction of axes (Translation)**: Let OX, OY be the original axes of coordinates. Let O'X', O'Y' be the new axes of coordinates parallel to the original axes through the new origin O'. Let (α, β) be the coordinates of O' referred to the original axes. Let P be any point whose coordinates referred to the old axes are (x, y) and referred to the new axes (x', y'). Draw PM parallel to OY, cutting OX in M and O'X' in M'. Draw O'N perpendicular to OX. Then OM = x, O'M' = x', PM = y, PM' = y', ON = a and O'N = β. From the geometry of the figure, we get, OM = ON + NM = ON + O'M' or, x = a + x' and PM = MM' + PM' = O'N + PM' or, y = β + y' .. x = x + α y = y + β (A) and x' = x - α y' = y - β (A) and (A') are the equations of transformation from the old to the new axes and from the new to the old respectively. **10. To transform from one set of rectangular axes system to another with the same origin (Rotation):** Let OX and OY be the original axes of coordinates, and OX', OY' be the new axes through the same origin and θ be the angle through which they have been turned in the same sense. That is ZXΟΧ' = ΖΥΟΥ' = θ. Let P be any point whose coordinates are (x, y) referred to the original axes and (x', y') referred to the new axes of coordinates. Draw PM and PM' respectively perpendiculars to OX and OX'. Draw M'N parallel to OY cutting OX in N and M'N' parallel to OX cutting PM in N'. Then OM=x, OM' = x', PM=y, PM'=y'. Again PM perpendicular to OX, and PM' perpendicular to OX' ZMPM' = ZXOX' = θ. Now OM = ON - MN = ON - M'N' = OM' cos 0 - PM' sin 0 or, x = x' cos 0 - y' sin 0 and PM = MN' + PN' = M'N + PN' = OM' sin 0 + PM' cos 0 or, y = x' sin 0 + y' cos 0. Thus x = x' cos 0 - y' sin 0 and y = x' sin 0 + y' cos 0 Solving the equations in (B), we get x' = x cos 0 + y sin 0 y' = -x sin θ + y cos 0 (B) (Β΄) These transforming equations (B) and (B') may be conveniently remembered from the following scheme: | | x | y | | -------- | - | - | | x' | cos θ | -sin θ | | y' | sin θ | cos θ | Fig. 12 Which may be read either horizontally or vertically. The transformed equation of a curve, in this transformation, is then obtained by substituting x cos 0 - y sin 0 for x and x sin 0 + y cos 0 for y. **Cor. If the origin is first transferred to the point (α, β) and the new set of axes through this origin be inclined at an angle θ to the original axes, the formulae of transformation will obviously be the combination of the above two sets of formulae. i.e.** x = a + x' cos 0 - y' sin 0 and y = β + x' sin 0 + y' cos 0 } (C) **Ex. Transform to axes inclined at 45° to the original axes the equation x² - y² = a².** Here θ = 45°.. cos θ = cos 45° = 1/√2 and sin θ = sin 45° = 1/√2 Then x = x' cos θ - y' sin θ = 1/√2 (x' - y') and y = x' sin θ + y' cos θ = 1/√2 (x' + y'). **11. Effect of transformation of coordinates upon the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0** [ This equation represents the general equation of the second degree ] **(a)** First consider the transformation x=x'+α and y=y'+β. When substituted in (1), the equation is transformed to a(x+α)² + 2h(x+α)(y+β) + b(y+β)² + 2g(x+α) + 2f(y+β) + c = 0 or, ax² + 2hxy + by² + 2(aa + hβ + g)x + 2(ha + bβ + f)y + aa² + 2haβ + bβ² + 2ga + 2fB + c = 0 [omitting accents] We see from (1) and (2) that the terms of the highest degree are unchanged and the constant term is the result of substituting a for x and ẞ for y in the original expression. If we write F(x, y) = ax² + 2hxy + by² + 2gx + 2fy + c, then F(α, β) = aa² + 2haβ + bβ² + 2ga + 2fẞ + c, δF/δa= 2aa + 2hβ + 2g = 2(aa + hβ + g), δF/δβ = 2ha + 2bβ + 2f = 2(ha + bβ + f). Substituțing these in (2), the transformed equation is cast in the simple form as ax² + 2hxy + by² + x δF/δa + y δF/δβ + F(α, β) = 0 Expressing this as a'x² + 2h'xy + b'y² + 2g'x + 2f'y + c' = 0 where a' = a, b' = b, h' = h, g' = δF/δa = aa + hβ + g, f' = δF/δβ = ha + bβ + f, c' = F(α, β) = πα² + 2hαβ + bβ² + 2ga + 2fẞ + c, we get (i) a' + b' = a + b (ii) a'b' - h² = ab - h² Also denoting the expression (abc + 2fgh - af² – bg² – ch²) by ∆, that is, letting Δ = abc + 2fgh - aff - bg2 - ch² = |- af² - ahg | | - hbf - gfc | we have, Δ' = |- af² - ahg | |- hbf - gfc | = |- a - aa + hβ + g | | - h - ha + bg +f | | - g - ga+fẞ+ c | = |- a - aa + hβ + g | | - h - ha + bβ + f | | - g - ga + fẞ + c | = | - a - aa + hβ + g | | - h αa + hβ + g | | - g - ga + fβ + c | [taking 1/3 = 13 - (a.r₁ + β.r2)] = | - a - ahg | | - h - hbf | | - g - gfc | [taking c'3 = C3 - (a.c₁ + β.c₂)] that is, (iii) Δ' = Δ or, a'b'c' + 2f'g'h' – a'f'2 – b'g'² - c'h'2 = abc + 2fgh - af² - bg2 - ch². **(b)** Now consider the transformation x= x' cos θ - y' sin θ y = x' sin θ + y' cos θ The transformed equation in this case is a(x cos θ - y sin θ)2 + 2lh (x cos θ - y sin θ) (x sin θ + y cos θ) + b(x sin ) + y cos θ)2 + 2g(x cos θ – y sin 0) + 2f(x sin θ + y cos θ) + c = 0 (4) [omitting accents] Let this be simplified to a'x² + 2h'xy + b'y² + 2g'x + 2f'y + c' = 0 (5) whence from a comparison of (4) and (5), we get a' = coefficient of x² = a cos2θ + 2l1 cos 0 sin 0 + b sin2θ = a (1+cos2θ) + h sin 20 + b (1-cos2θ)/2 = a+b / 2 + (a-b)/2 * cos 2θ + h sin 2θ b' = coefficient of y² = a sin2θ - 2l1 cose sine + b cos2θ = a (1- cos 2θ)/2 - li sin 2θ + b (1 + cos 2θ)/2 = a+b / 2 + (a-b)/2 * cos 2θ - li sin 2θ and 2li' = co-efficient of xy = 2h(cos² 0 - sin² 0) – 2a cos 0 sin 0 + 2b cos 0 sin 0 = 2h cos 2θ – (a – b) sin 2θ or, li' = h cos 2θ – (a – b)/2 * sin 2θ g' = coefficient of 2x = g cos 0 + f sin 0, f' = coefficient of 2y = - g sin 0 + f cos 0, c' = c From (6), (7) and (8), a' + b' = + (a+b)/2 * cos 2θ + li sin 2θ + (a+b)/2 + (a-b)/2 * cos 2θ + li sin 2θ) = a + b i.e., (i) a' + b' = a + b and a'b' - h'2 = (((a+b)/2 + (a-b)/2 * cos 2θ + h sin 2θ)) (((a+b)/2 + (a-b)/2 * cos 2θ - h sin 2θ)) - (h cos 2θ – (a – b)/2 * sin 2θ)² = (((a+b)/2)² - ((a-b)/2)² * cos² 2θ - h² * sin² 2θ) - ( h² * cos² 2θ - (a-b)h/2 * sin 2θ*cos 2θ + (a-b)h/2 * sin 2θ*cos 2θ - ((a-b)/2)² * sin² 2θ) = (((a+b)/2)² - ((a-b)/2)² * cos² 2θ - h² * sin² 2θ) - ( h² * cos² 2θ - ((a-b)/2)² * sin² 2θ) = (a+b)²/4 - (a-b)²/4 * cos² 2θ h² * sin² 2θ - h² * cos² 2θ + (a-b)²/4 * sin² 2θ) = (a+b)²/4 - (a-b)²/4 - h² = (a+b)/2 * (a-b)/2 - h² = ab - h² or, (ii) a'b' - h′2 = ab – h². **Cor. If we require that the term xy will be absent in the transformed equation, then h' = 0.** Hence from (8), h cos 2θ - (a-b)/2 * sin 2θ = 0 or, tan 2θ = 2h/(a-b) i.e., θ = 1/2 * tan-1 (2h/(a-b)) (10) Therefore, if the axes are rotated through an angle θ = 1/2 * tan-1 (2h/(a-b)) the term xy will be absent in the transformed equation. Again Δ' = a'b'c' + 2f'g'h' - a'f'2 - b'g'2 - c'h'2 = c'(a'b' - h′2) + 2( - g sin 0 + f cos 0) (g cos 0 + f sin 0)h' - a'( - g sin 0 + f cos 0)2 - b'(g cos 0 + f sin 0)2 or, Δ' = c(ab – h²) + [(f2 – g2) sin 20 + 2fg cos 20] lh' = c(ab - h²) - (a² +b') + (f² + g²)/2 -a' * (f² - g²)/2 * cos 20 - fg * sin 20 -b' * (f² - g²)/2 * cos 20 - fg * sin 20 + [(f2g2) sin 20 + 2fg cos 20)]h' = c(ab - h²) - (a + b) + (f² + g²)/2 - (f² - g²)/2 * cos 20 - fg * sin 20 * (d' - b') = c(ab - h²) - (a + b) + (f² + g²)/2 - (f² - g²)/2 * cos 20 - fg * sin 20 * (d' - b') + [(f2 - g²) sin 20 + 2fg cos 20]. h cos 20 - (a-b)/2 * sin 20 - [(f2 - g²) cos 20 - 2fg sin 20] li sin 20 + (a-b)/2 * cos 20 = c(ab-h²)_+ (f2+82)-(2-82) + 2fgh = abc + 2fgh - af2 – bg2 – ch² = A .. (iii) Δ' = Δ. **(c)** **Invariants:** The discussions raveal that if there is a translation or rotation of the axes of coordinates or a combination of both, the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 transforms to a'x² + 2h'xy + b'y² + 2g'x + 2f'y + c' = 0 in which (i) a' + b' = a + b √ (ii) a'b' - h^2 = ab - h² √ (iii) A' = A or, a'b'c' + 2f'g'h' – a'f2 – b'g'² - c'h² = abc + 2fgh - af² - bg2 - ch². The quantities like a + b, ab – h² and A which remain unaltered on all possible transformations are the invariants. **Ex. Verify that when the axes are turned through an angle π/4, the equation 5x2 + 4xy + 5y2 - 10 = 0 transforms to one in which the term xy is absent.** Here a = 5, b = 5 and h = 2 If the axes be turned through an angle θ = 1/2 * tan-1 (2h/(a-b)) = 1/2 * tan-1 (2/1) = 1 / 2 * tan-1 ∞ the term xy will be absent in the transformed equation. Again, if the axes are rotated through π/4, the transformed equation is obtained by substituting x= x cos π/4 - y sin π/4 = 1/√2 (x - y) and y = x sin π/4 + y cos π/4 = 1/√2 (x + y) in the original equation. Thus the transformed equation is 5(1/√2 (x - y))² + 4(1/√2 (x - y))(1/√2 (x + y)) + 5(1/√2 (x + y))² - 10 = 0 or, 5(x² - 2xy + y²)/2 + 4(x² - y²)/2 + 5(x² + 2xy + y²)/2 - 10 = 0 or, 5(x²-2xy + y²) + 4(x² - y²) + 5(x2 + 2xy + y²) — 20 = 0 or, 14x2 + 6y2 - 20 = 0 or, 7x2 + 3y²- 10 = 0, in which the term xy is absent. Hence the result. ## Illustrative Examples **Ex. Transform the equation 9x2 + 24xy + 2y2 - 6x + 20y + 41 = 0 in rectangular coordinates so as to remove the terms in x, y and xy.** Let us first transform the equation to parallel axes through the point (α, β). The transformed equation is then 9(x'+ α)² + 24(x + α) (y + β) + 2(y + β)² - 6(x + a) + 20(y + β) + 41 = 0 or, 9x + 24xy + 2y² + (18α + 24β – 6)x + (24a + 43 + 20)y +9α2 + 24αβ + 2β2 – 6α + 20β + 41= 0 The terms in x and y in (1) will be absent if 18α + 24β − 6 = 0, and 24a + 4ẞ + 20 = 0 ; that is, if a = - 1, and ẞ = 1. Hence when the given equation is transformed to parallel axes through the point (-1, 1), it becomes 9x2 + 24xy + 2y² + 54 = 0 Now in order to remove the term in xy, let the axes be rotated throguh the angle θ. Therefore, replacing x by (x cos 0 y sin 0) and y by (x sin 0 + y cos 0), we have from (2), 9(x cos 0 - y sin 0)2 + 24(x cos 0 – y sin 0) (x sin 0 + y cos 0) + 2(x sin 0 + y cos 0)2 + 54 = 0 or, (9 cos² 0 + 24 cos 0 sin 0 + 2 sin2 0)x2 +\2{ − 7 cos 0 sin 0 + 12(cos² 0 - sin² 0)} xy + (9 sin20 -24 sin 0 cos 0 + 2 cos2 0)y2 = 0 or, (9 cos² 0 + 24 cos 0 sin 0 + 2 sin2 0)x2 + 2(- 7/2 * sin 2θ + 12 cos2θ) xy + (9 sin20 -24 sin 0 cos 0 + 2 cos2 0)y2 = 0 by our assumption, -7 cos 0 sin 0 + 12(cos² 0 - sin² 0) = 0 -7/2 * sin 2θ + 12 cos 20 = 0 i.e., tan 20 = 24/7 .. 20 = tan-124 = 2 tan-1 3/4 or, θ = tan-1 3/4 So, sin 0 = 3/5 and cos 0 = 4/5 ## Exercise 2 Transform the following equations : (Both sets of axes being rectangular). 1. 3x - 25y + 41 = 0 to parallel axes through (– 3, 2). √ 2. x² + y² - 8x + 14y + 5 = 0 to parallel axes through (4, — 7). √ 3. y(y - 2a) = ax to parallel axes through ( — а, а). 4. 3x² + 5y² - 3 = 0 to axes turned through 45°. 5. 7x² - 2xy + y² + x + y + 17 = 0 to axes through the angle tan-1 1/2. 6. 2x - 2xy + 9y² + 1 = 0 to axes turned through the point (-1, 2) inclined at an angle tan-1 1/4 to the original axes. 7. 11x² + 3xy + 7y² + 19 = 0 so as to remove the term in xy. 8. 9x² + 15xy + y² + 12x - 11y - 15 = 0 so as to remove the terms in x, y and xy. √ 9. (ax + by + c) (bx - ay + d) = a² + b² to axes through the point (-(ac + bd)/(a² + b²), (ad-bc)/(a² + b²)) inclined at an angle tan-1(b/a) to the original axes. √ 10. x² - 2ax + 2by + c² = 0 to axes through the point (a, b) inclined at an angle π/4 to the original axes. 11. By transforming to parallel axes through a properly chosen point (h, k), prove that the equation 3x² – 5xy + y² + 7x + 5y - 23 = 0 can be reduced to one containing only terms of the second degree. ## Answers 1. 3x - 25y - 18 = 0. 2. x² + y² - 60 = 0. 3. y² = ax 4. 4x² + 2xy + 4y² - 3 = 0. 5. 5x² - 6xy + 3y² + 1 = 0. 6. 89x² + 154xy + 186y² - 25x + 175y + 1550 = 0. 7. To axes rotated through the angle tan-1 1/3: 23x² + 13y² + 38 = 0. 8. 27x²-7y² + 4 = 0. 9. xy = 1. 10. 2xy = b² + c² - a². 11. (3,5).