{"id":10659,"date":"2024-02-29T10:39:01","date_gmt":"2024-02-29T05:09:01","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10659"},"modified":"2024-03-05T12:53:30","modified_gmt":"2024-03-05T07:23:30","slug":"maths-1b-transformation-of-axes-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/maths-1b-transformation-of-axes-important-questions\/","title":{"rendered":"TS Inter First Year Maths 1B Transformation of Axes Important Questions"},"content":{"rendered":"

Students must practice these Maths 1B Important Questions<\/a> TS Inter First Year Maths 1B Transformation of Axes Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter First Year Maths 1B Transformation of Axes Important Questions<\/h2>\n

Question 1.
\nWhen the origin is shifted to (3, 4) by the translation of axes, find the transformed equation of 2x2<\/sup> + 4xy + 5y2<\/sup> = 0. [Mar. ’00]
\nSolution:
\nGiven equation is 2x2<\/sup> + 4xy + 5y2<\/sup> = 0 ……..(1)
\nLet (X, Y) be the new coordinates of the point (x, y).
\nGiven a point, (h, k) = (3, 4)
\nBy the translation of axes,
\nx = X + h, y = Y + k
\nx = X + 3, y = Y + 4
\nThe transformed equation of (1) is,
\n2(X + 3)2<\/sup> + 4(X + 3)(Y + 4) + 5(Y + 4)2<\/sup> = 0
\n\u21d2 2(X2<\/sup> + 6X + 9) + 4(XY + 4X + 3Y + 12) + 5(Y2<\/sup> + 8Y + 16) = 0
\n\u21d2 2X2<\/sup> + 12X + 18 + 4XY + 16X + 12Y + 48 + 5Y2<\/sup> + 40Y + 80 = 0
\n\u21d2 2X2<\/sup> + 5Y2<\/sup> + 28X + 52Y – 4XY + 146 = 0
\n\u21d2 2X2<\/sup> + 4XY + 5Y2<\/sup> + 28X + 52Y + 146 = 0<\/p>\n

Question 2.
\nWhen the origin is shifted to the point (2, 3), the transformed equation of a curve is x2<\/sup> + 3xy – 2y2<\/sup> + 17x – 7y – 11 = 0. Find the original equation of the curve. [Mar. ’16 (TS), ’13, ’11, ’10, ’09; May ’15 (AP), ’09]
\nSolution:
\nThe given equation is,
\nX2<\/sup> + 3XY – 2Y2<\/sup> + 17X – 7Y – 11 = 0 ………(1)
\nLet (x, y) be the original coordinates of the point (X, Y).
\nGiven a point, (h, k) = (2, 3).
\nBy the translation of axes,
\nx = X + h, y = Y + k
\n\u21d2 x = X + 2, y = Y + 3
\n\u21d2 X = x – 2, Y = y – 3
\nThe original equation of (1) is
\n(x – 2)2<\/sup> + 3(x – 2)(y – 3) – 2(y – 3)2<\/sup> + 17(x – 2) – 7(y – 3) – 11 = 0
\n\u21d2 x2<\/sup> + 4 – 4x + 3(xy – 3x – 2y + 6) – 2(y2<\/sup> + 9 – 6y) + 17x – 34 – 7y + 21 – 11 = 0
\n\u21d2 x2<\/sup> + 4 – 4x + 3xy – 9x – 6y + 18 – 2y2<\/sup> – 18 + 12y + 17x – 34 – 7y + 21 – 11 = 0
\n\u21d2 x2<\/sup> + 3xy – 2y2<\/sup> + 4x – y – 20 = 0
\n\u2234 The original equation is x2<\/sup> + 3xy – 2y2<\/sup> + 4x – y – 20 = 0<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the point to which the origin is to be shifted so as to remove the first-degree terms from the equation 4x2<\/sup> + 9y2<\/sup> – 8x + 36y + 4 = 0. [May ’03]
\nSolution:
\nGiven equation is 4x2<\/sup> + 9y2<\/sup> – 8x + 36y + 4 = 0
\nComparing the given equation with ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0
\nWe get a = 4, h = 0, b = 9, g = -4, f = 18, c = 4.
\n\"TS<\/p>\n

Question 4.
\nFind the point to which the origin is to be shifted by the translation of axes 50 as to remove the first-degree terms from the equation ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0 where h2<\/sup> \u2260 ab. [May ’97]
\nSolution:
\nGiven equation is ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0 …….(1)
\nLet (\u03b1, \u03b2) be the required point.
\nLet (X, Y) be the new coordinates of the point (x, y).
\n\u2234 x = X + \u03b1, y = Y + \u03b2
\nNow substitute the values of x, y in equation (1)
\n\u2234 The transformed equation is a(X + \u03b1)2<\/sup> + 2h(X + \u03b1)(Y + \u03b2) + b(Y + \u03b2)2<\/sup> + 2g(X + \u03b1) + 2f(Y + \u03b2) + c = 0
\n\u21d2 a(X2<\/sup> + \u03b12<\/sup> + 2X\u03b1) + 2h(XY + \u03b2X + \u03b1Y + \u03b1\u03b2) + b(Y2<\/sup> + \u03b22<\/sup> + 2\u03b2Y) + 2gX + 2g\u03b1 + 2fY + 2f\u03b2 + c = 0
\n\u21d2 aX2<\/sup> + a\u03b12<\/sup> + 2aX\u03b1 + 2hXY + 2h\u03b2X + 2h\u03b1Y + 2h\u03b1\u03b2 + bY2<\/sup> + b\u03b22<\/sup> + 2b\u03b2Y + 2gX + 2g\u03b1 + 2fy + 2f\u03b2 + c = 0
\n\u21d2 aX2<\/sup> + 2hXY + bY2<\/sup> + 2X(a\u03b1 + h\u03b2 + g) + 2Y(h\u03b1 + b\u03b2 + f) + a\u03b12<\/sup> + 2h\u03b1\u03b2 + b\u03b22<\/sup> + 2g\u03b1 + 2f\u03b2 + c = 0
\nSince, x, y terms are eliminated
\na\u03b1 + h\u03b2 + g = 0 ……..(2)
\nh\u03b1 + b\u03b2 + f = 0 ……….(3)
\nSolving (2) & (3)
\n\"TS<\/p>\n

Question 5.
\nProve that the angle of rotation of the axes to eliminate xy term from the equation ax2<\/sup> + 2hxy + by2<\/sup> = 0 is \\(\\frac{1}{2} \\tan ^{-1}\\left(\\frac{2 h}{a-b}\\right)\\) where a \u2260 b and \\(\\frac{\\pi}{4}\\) if a = b. [Mar. ’13 (Old); ’13, ’06, ’02]
\nSolution:
\nx = X cos \u03b8 – Y sin \u03b8
\ny = X sin \u03b8 + Y cos \u03b8
\nGiven equation is ax2<\/sup> + 2hxy + by2<\/sup> = 0 ……..(1)
\n\u2234 The transformed equation of (1) is
\na(X cos \u03b8 – Y sin \u03b8)2<\/sup> + 2h(X cos \u03b8 – Y sin \u03b8) (X sin \u03b8 + Y cos \u03b8) + b(X sin \u03b8 + Y cos \u03b8)2<\/sup> = 0
\n\u21d2 a[X2<\/sup> cos2<\/sup>\u03b8 + Y2 sin2<\/sup>\u03b8 – 2X cos \u03b8 . Y sin \u03b8] + 2h(X2<\/sup> cos \u03b8 . sin \u03b8 + XY cos2<\/sup>\u03b8 – XY sin2<\/sup>\u03b8 + Y2<\/sup> sin \u03b8 cos \u03b8 + b[X2<\/sup> sin2<\/sup>\u03b8 + Y2<\/sup> cos2<\/sup>\u03b8 + 2XY sin \u03b8 cos \u03b8] = 0
\n\u21d2 [aX2<\/sup> cos2<\/sup>\u03b8 + aY2<\/sup> sin2<\/sup>\u03b8 – 2aX cos \u03b8 . Y sin \u03b8 + 2hX2<\/sup> cos \u03b8 . sin \u03b8 + 2hXY cos2<\/sup>\u03b8 – 2hXY sin2<\/sup>\u03b8 + 2hY2<\/sup> sin \u03b8 cos \u03b8 + bX2<\/sup> sin2<\/sup>\u03b8 + bY2<\/sup> cos2<\/sup>\u03b8 + 2bXY sin \u03b8 cos \u03b8] = 0
\n\u21d2 X2<\/sup>[a cos2<\/sup>\u03b8 + 2h cos \u03b8 . sin \u03b8 + b sin2<\/sup>\u03b8) + XY (-2a sin \u03b8 cos \u03b8 + 2h cos2<\/sup>\u03b8 – 2h sin2<\/sup>\u03b8 + 2b sin \u03b8 cos \u03b8) + Y2<\/sup> (a sin2<\/sup>\u03b8 – 2h sin \u03b8 cos \u03b8 + b cos2<\/sup>\u03b8] = 0
\nSince xy term is eliminated, then,
\n\u21d2 -2a sin \u03b8 . cos \u03b8 + 2h cos2<\/sup>\u03b8 – 2h sin2<\/sup>\u03b8 + 2b sin \u03b8 cos \u03b8 = 0
\n\u21d2 -a sin 2\u03b8 + 2h(cos2<\/sup>\u03b8 – sin2<\/sup>\u03b8) + b sin 2\u03b8 = 0
\n\u21d2 -a sin 2\u03b8 + 2h cos 2\u03b8 + b sin 2\u03b8 = 0
\n\u21d2 2h cos 2\u03b8 = a sin 2\u03b8 – b sin 2\u03b8
\n\u21d2 2h cos 2\u03b8 = (a – b) (sin 2\u03b8)
\n\"TS<\/p>\n

Question 6.
\nFind the angle through which the axes are to be rotated so as to remove the xy term in the equation x2<\/sup> + 4xy + y2<\/sup> – 2x + 2y – 6 – 0. [May ’04, ’96]
\nSolution:
\nGiven equation is x2<\/sup> + 4xy + y2<\/sup> – 2x + 2y – 6 = 0
\nBy comparing the given equation with ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0,
\nwe get a = 1, h = 2, b = 1
\nThe required angle of rotation
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 7.
\nWhen the axes are rotated through an angle of 60\u00b0, the new coordinates of the point are (3, 4). Find their original coordinates. [May ’03]
\nSolution:
\nLet (x, y) be the original coordinates of (X, Y).
\nGiven that, the angle of rotation \u03b8 = 60\u00b0
\n(X, Y) = (3, 4) then
\nx = X cos \u03b8 – Y sin \u03b8
\n= 3(cos 60\u00b0) – 4(sin 60\u00b0)
\n\"TS<\/p>\n

Question 8.
\nWhen the axes are rotated through an angle of 45\u00b0, the transformed equation of a curve is 17x2<\/sup> – 16xy + 17y2<\/sup> = 225. Find the original equation of the curve. [May ’15 (TS), ’12, ’10, ’01; Mar. ’15 (TS), ’08]
\nSolution:
\nGiven equation is 17x2<\/sup> – 16xy + 17y2<\/sup> = 225 ………(1)
\nAngle of rotation \u03b8 = 45\u00b0
\nLet (x, y) be the original coordinates of (X, Y) then
\nX = x cos \u03b8 + y sin \u03b8
\n= x cos 45\u00b0 + y sin 45\u00b0
\n\"TS
\n50x2<\/sup> + 18y2<\/sup> = 450
\n25x2<\/sup> + 9y2<\/sup> = 225<\/p>\n

Question 9.
\nWhen the axes are rotated through an angle \u03b1, find the transformed equation of x cos \u03b1 + y sin \u03b1 = P. [Mar. ’19 (TS); Mar. ’14, ’00; May ’07]
\nSolution:
\nGiven equation is x cos \u03b1 + y sin \u03b1 = P ………..(1)
\nLet (X, Y) be the new coordinates of (x, y)
\nGiven that, the angle of rotation \u03b8 = \u03b1, then
\nx = X cos \u03b8 + Y sin \u03b8 = X cos \u03b1 + Y sin \u03b1
\ny = X sin \u03b8 + Y cos \u03b8 = X sin \u03b1 + Y cos \u03b1
\nThe transformed equation of (1) is
\n(X cos \u03b1 – Y sin \u03b1) cos \u03b1 + (X sin \u03b1 + Y cos \u03b1) sin \u03b1 = P
\n\u21d2 X cos2\u03b1 – Y sin \u03b1 cos \u03b1 + X sin2\u03b1 + Y cos \u03b1 sin \u03b1 = P
\n\u21d2 X cos2<\/sup>\u03b1 + X sin2<\/sup>\u03b1 = P
\n\u21d2 X(cos2<\/sup>\u03b1 + sin2<\/sup>\u03b1) = P
\n\u21d2 X(1) = P
\n\u21d2 X = P<\/p>\n

Question 10.
\nWhen the axes are rotated through an angle \\(\\frac{\\pi}{6}\\), find the transformed equation of x2<\/sup> + 2\u221axy – y2<\/sup> = 2a2<\/sup>. [Mar. ’15 (AP); May ’13, ’06, ’03; Mar. ’12, ’07, ’03; B.P.; Mar. ’18 (TS)]
\nSolution:
\nGiven equation is x2<\/sup> + 2\u221a3 xy – y2<\/sup> = 2a2<\/sup> ………(1)
\nLet (X, Y) be the new coordinate of (x, y).
\nGiven that, angle of rotation \u03b8 = \\(\\frac{\\pi}{6}\\) = 30\u00b0, then
\nx = X cos \u03b8 – Y sin \u03b8
\n= X cos 30\u00b0 – Y sin 30\u00b0
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 11.
\nWhen the axes are rotated through an angle \\(\\frac{\\pi}{4}\\), find the transformed equation 3x2<\/sup> + 10xy + 3y2<\/sup> = 9. [Mar. (AP & TS) ’17; May ’14, ’11, ’05; Mar. ’08]
\nSolution:
\nGiven equation is, 3x2<\/sup> + 10xy + 3y2<\/sup> = 9 ……(1)
\nLet (X, Y) be the new coordinates of (x, y).
\nGiven that, angle of rotation \u03b8 = \\(\\frac{\\pi}{4}\\) = 45\u00b0
\nx = X cos \u03b8 – Y sin \u03b8
\n= X cos 45\u00b0 – Y sin 45\u00b0
\n\"TS<\/p>\n

Some More Maths 1B Transformation of Axes Important Questions<\/h3>\n

Question 12.
\nWhen the origin is shifted to (-1, 2) by the translation of axes find the transformed equation of x2<\/sup> + y2<\/sup> + 2x – 4y + 1 = 0.
\nSolution:
\nGiven equation is x2<\/sup> + y2<\/sup> + 2x – 4y + 1 = 0 ……(1)
\nLet (X, Y) be the new coordinates of the point (x, y).
\nGiven that, (h, k) = (-1, 2)
\nBy the translation of axes,
\nx = X + h, y = Y + k
\n\u21d2 x = X- 1, y = Y + 2
\nThe transformed equation of (1) is
\n(X – 1)2<\/sup> + (Y + 2)2<\/sup> + 2(X – 2) – 4(Y + 2) + 1 = 0
\n\u21d2 X2<\/sup> – 2X + 1 + Y2<\/sup> + 4 + 4Y + 2X – 2 – 4Y – 8 + 1 = 0
\n\u21d2 X2<\/sup> + Y2<\/sup> – 4 = 0
\n\u2234 The transformed equation is X2<\/sup> + Y2<\/sup> – 4 = 0<\/p>\n

Question 13.
\nWhen the origin is shifted to (-1, 2) by the translation of axes, find the transformed equation of 2x2<\/sup> + y2<\/sup> – 4x + 4y = 0.
\nSolution:
\nGiven equation is 2x2<\/sup> + y2<\/sup> – 4x + 4y = 0 …..(1)
\nLet (X, Y) be the new coordinates of the point (x, y).
\nGiven that, (h, k) = (-1, 2)
\nBy the translation of axes,
\nx = X + h, y = Y + k
\n\u21d2 x = X – 1, y = Y + 2
\nThe transformed equation of (1) is,
\n2(X – 1)2<\/sup> + (Y + 2)2<\/sup> – 4(X – 1) + 4(Y + 2) = 0
\n\u21d2 2(X2<\/sup> – 2X + 1) + (Y2<\/sup> + 4 + 4Y) – 4X + 4 + 4Y + 8 = 0
\n\u21d2 2X2<\/sup> – 4X + 2 + Y2<\/sup> + 4 + 4Y – 4X + 4 + 4Y + 8 = 0
\n\u21d2 2X2<\/sup> + Y2<\/sup> – 8X + 8Y + 18 = 0
\n\u2234 The transformed equation is 2X2<\/sup> + Y2<\/sup> – 8X + 8Y + 18 = 0<\/p>\n

\"TS<\/p>\n

Question 14.
\nWhen the origin is shifted to the point (3, -4), the transformed equation of a curve is x2<\/sup> + y2<\/sup> = 4. Find the original equation of the curve.
\nSolution:
\nGiven equation is X2<\/sup> + Y2<\/sup> = 4 ………(1)
\nLet (x, y) be the original coordinates of the point (X, Y).
\nGiven point, (h, k) = (3, -4)
\nBy the translation of axes,
\nx = X + h, y = Y + k
\n\u21d2 x = X + 3, y = Y – 4
\n\u21d2 X = x – 3, Y = y + 4
\nThe original equation of (1) is (x – 3)2<\/sup> + (y + 4)2<\/sup> = 4
\n\u21d2 x2<\/sup> + 9 – 6x + y2<\/sup> + 16 + 8y = 4
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x + 8y + 25 = 4
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x + 8y + 25 – 4 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x + 8y + 21 = 0
\n\u2234 The original equation is x2<\/sup> + y2<\/sup> – 6x + 8y + 21 = 0.<\/p>\n

Question 15.
\nWhen the origin is shifted to the point (-1, 2), the transformed equation of a curve is x2<\/sup> + 2y2<\/sup> + 16 = 0. Find the original equation of the curve.
\nSolution:
\nGiven equation is X2<\/sup> + 2Y2<\/sup> + 16 = 0 …….(1)
\nLet (x, y) be the original coordinates of the point (X, Y).
\nGiven a point, (h, k) = (-1, 2)
\nBy the translation of axes,
\nx = X + h, y = Y + k
\n\u21d2 x = X – 1, y = Y + 2
\n\u21d2 X = x + 1, Y = y – 2
\nThe original equation of (1) is (x + 1)2<\/sup> + 2(y – 2)2<\/sup> + 16 = 0
\n\u21d2 x2<\/sup> + 2x + 1 + 2(y2<\/sup> + 4 – 4y) + 16 = 0
\n\u21d2 x2<\/sup> + 2x + 1 + 2y2<\/sup> + 8 – 8y + 16 = 0
\n\u21d2 x2<\/sup> + 2y2<\/sup> + 2x – 8y + 25 = 0
\n\u2234 The original equation is x2<\/sup> + 2y2<\/sup> – 2x – 8y + 25 = 0.<\/p>\n

\"TS<\/p>\n

Question 16.
\nWhen the origin is shifted to (-2, -3) and the axes are rotated through an angle of 45\u00b0, find the transformed equation of 2x2<\/sup> + 4xy – 5y2<\/sup> + 20x – 22y – 14 = 0.
\nSolution:
\nGiven equation is 2x2<\/sup> + 4xy – 5y2<\/sup> + 20x – 22y – 14 = 0 ………(1)
\nLet (X, Y) be the new coordinates of (x, y).
\nGiven that, (h, k) = (-2, -3)
\nangle of rotation \u03b8 = 45\u00b0
\nx = X cos \u03b8 – Y sin \u03b8 + h
\n= X cos 45 – Y sin 45 + (-2)
\n\"TS
\n\"TS
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Transformation of Axes Important Questions to help strengthen their preparations for exams. TS Inter First Year Maths 1B Transformation of Axes Important Questions Question 1. When the origin is shifted to (3, 4) by the translation of axes, find the transformed … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10659"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10659"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10659\/revisions"}],"predecessor-version":[{"id":10672,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10659\/revisions\/10672"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10659"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10659"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}