{"id":35141,"date":"2022-08-25T10:51:35","date_gmt":"2022-08-25T05:21:35","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35141"},"modified":"2022-08-25T10:51:35","modified_gmt":"2022-08-25T05:21:35","slug":"inter-1st-year-maths-1b-transformation-of-axes-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-transformation-of-axes-important-questions\/","title":{"rendered":"Inter 1st Year Maths 1B Transformation of Axes Important Questions"},"content":{"rendered":"

Students get through Maths 1B Important Questions<\/a> Inter 1st Year Maths 1B Transformation of Axes Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 1st Year Maths 1B Transformation of Axes Important Questions<\/h2>\n

Question 1.
\nWhen the origin is shifted to (2, 3) by the translation of axes, the coordinates of a point p are changed as (4, -3). Find the coordinates of P in the original system.
\nSolution:
\n(h, k) = (2, 3) \u21d2 h = 2, k = 3
\n(x’, y’) = (4, 3) \u21d2 x = 4, y = -3
\nx = x’ + h = 4 + 2 = 6, y = y’ + k = -3 + 3 = 0
\noriginal Co-ordinates are (6, 0)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
\nax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0, where h2<\/sup> \u2260 ab.
\nSolution:
\nLet the origin be shifted to (\u03b1, \u03b2) by the translation of axes.
\nThen x = x’ + \u03b1,
\ny = y’ + \u03b2
\nOn substituting these in the given equation, we get
\na(x’ + \u03b1)2<\/sup> + 2h(x’ + \u03b1) (y’ + \u03b2) + b(y’ + \u03b2)2<\/sup> + 2g(x’ + \u03b1) + 2f(y’ + \u03b2) + c = 0
\nWhich gives
\nax’2<\/sup> + 2hx’y’ + by2<\/sup> + 2x'(\u03b1 + \u03b2 + g) + 2y'(h\u03b1 + b\u03b2 + f) + a\u03b12<\/sup> + 2h\u03b1\u03b2 + b\u03b22<\/sup> + 2g\u03b1 + 2f\u03b2 + c = 0 ……….. (1)
\nIf equation (1) has to be free from the first degree terms, then we have
\na\u03b1 + h\u03b2 + g = 0 and h\u03b1 + b\u03b2 +1 = 0
\nSolving these equations for a and 13, we get
\n\u03b1 =\\(\\frac{h f-b g}{a b-h^{2}}\\) , \u03b2 = \\(\\frac{g h-a f}{a b-h^{2}}\\)
\nTherefore, the origin is to be shifted to
\n\\(\\left(\\frac{h f-b g}{a b-h^{2}}, \\frac{g h-a f}{a b-h^{2}}\\right)\\)<\/p>\n

Question 3.
\nFind the point to which the origin is to be shifted by the translation of axes so as to remove the first degree terms from the equation
\nax2<\/sup> + by2<\/sup> + 2gx + 2fy + c = 0, where a \u2260 0, b \u2260 0.
\nSolution:
\nHere the given equation does not contain xy term. Hence writing h = 0 in the result of
\nProblem 12, the required point is \\(\\left(\\frac{-g}{a}, \\frac{-f}{b}\\right)\\)<\/p>\n

\"Inter<\/p>\n

Question 4.
\nIf the point P changes to (4, -3) when the axes are rotated through an angle of 135\u00b0, find the coordinates of P with respect to the original system.
\nSolution:
\nHere (x’, y’) = (4, -3); \u03b8 = 135\u00b0
\nLet (x, y) be the coordinates of P, then
\nx = x’ cos \u03b8 – y’ sin \u03b8
\n= 4cos 135\u00b0 – (-3) sin 135\u00b0
\n\"Inter
\nTherefore, the coordinates of P with respect to the original system are (\\(\\frac{-1}{\\sqrt{2}}\\), \\(\\frac{7}{\\sqrt{2}}\\))<\/p>\n

Question 5.
\nShow that the axes are to be rotated through an angle of \\(\\frac{1}{2}\\) Tan-1<\/sup>(\\(\\frac{2 h}{a-b}\\)) so as to remove the xy term from the equation ax2<\/sup> + 2hxy + by2<\/sup> = 0, if a \u2260 b and through the angle \\(\\frac{\\pi}{4}\\), if a = b. [Mar 13, 06]
\nSolution:
\nIf the axes are rotated through an angle \u2018\u03b8\u2019,
\nthen
\nx = x’ cos \u03b8 – y’ sin \u03b8
\ny = x’ sin \u03b8 + y’ cos \u03b8
\nTherefore the given equation transforms as
\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> = o
\nTo remove x’y’ term from the equation, the coefficient of x’y’ term must be zero.
\nSo, (b – a) sin \u03b8 cos \u03b8 + h(cos2<\/sup>\u03b8 – sin2<\/sup>\u03b8) = 0
\ni.e., h cos 2\u03b8 = \\(\\frac{a-b}{2}\\) sin 2\u03b8
\ni.e., tan 2\u03b8 = \\(\\frac{2 h}{a-b}\\), if a \u2260 b and h cos 2\u03b8 = 0, if a = b
\nTherefore \u03b8 = \\(\\frac{1}{2}\\) Tan-1<\/sup>(\\(\\frac{2 h}{a-b}\\)), if a \u2260 b and \u03b8 = \\(\\frac{\\pi}{4}\\), If a = b<\/p>\n

\"Inter<\/p>\n

Question 6.
\nWhen the origin is shifted to (-2, -3) and the axes are rotated through an angle 45\u00b0 find the transformed equation of 2x2<\/sup> + 4xy – 5y2<\/sup> + 20x – 14 = 0.
\nSolution:
\nHere (h, k) = (-2, -3), h = -2, k = -3
\n\u03b8 = 45\u00b0
\nLet (x’, y’) be the new co-ordinates of any point (x, y) is the plane after transformation
\nx = x’ cos \u03b8 – y’ sin \u03b8 + h = -2 + x’ cos 45\u00b0 – y’ sin 45\u00b0.
\n= -2 + \\(\\frac{x^{\\prime}-y^{\\prime}}{\\sqrt{2}}\\)
\ny = x’ si n\u03b8 + y’ cos \u03b8 + k = x’ sin 45\u00b0 + y’ cos
\n45\u00b0 – 3 = -3 + \\(\\frac{x^{\\prime}+y^{\\prime}}{\\sqrt{2}}\\)
\nThe transformed equation is
\n\"Inter
\n\u21d2 (x’ – y’)2<\/sup> + 8 – 4\\(\\sqrt{2}\\) (x’ – y’) + 2(x’2<\/sup> – y’2<\/sup>) – 6\\(\\sqrt{2}\\) (x’ – y’) – 4\\(\\sqrt{2}\\) (x’ + y’) + 24 = 0
\n\u21d2 –\\(\\frac{5}{2}\\) (x’ – y’)2<\/sup> – 45 + 15\\(\\sqrt{2}\\) (x’ + y’) + 10\\(\\sqrt{2}\\) (x’ – y’) – 40 – 11\\(\\sqrt{2}\\) (x’ + y’) + 66 – 14 = 0
\n\u21d2 x’2<\/sup> + y’2<\/sup> – 2x’y’ + 2x’2<\/sup> – 2y’2<\/sup> – \\(\\frac{5}{2}\\) (x’2<\/sup> + y’2<\/sup> + 2x’y’) – 1 = 0
\n\\(\\frac{1}{2}\\) x’2<\/sup> – \\(\\frac{7}{2}\\) y’2<\/sup> – 7x’y’ – 1 = 0
\ni.e., x’2<\/sup> – 7y’2<\/sup> – 14x’y’ – 2 = 0
\nThe transformed equation is (dropping dashes)
\nx2<\/sup> – 7y2<\/sup> – 14xy – 2 = 0<\/p>\n

Question 7.
\nWhen the origin is shifted to (-2, 3) by translation of axes, let us find the co-ordinates of (1, 2) with respect to new axes.
\nSolution:
\nHere (h, k) = (-2, 3)
\nLet (x, y) = (1, 2) be shifted to (x’ y’) by the translation of axes.
\nThen (x’, y’) = (x – h, y – k)
\n= (1 – (-2), 2 – 3) = (3, -1)<\/p>\n

\"Inter<\/p>\n

Question 8.
\nWhen the origin is shifted to (3, 4) by the translation of axes, let us find the transformed equation of 2x2<\/sup> + 4xy + 5y2<\/sup> = 0.
\nSolution:
\nHere (h, k) = (3, 4)
\non substituting x = x’ + 3 and y = y’ + 4 in the given equation.
\nas per note the equation f(x, y) = 0 of the curve is transformed as f(x’ + h, y’ + k) = 0]
\nwe get
\n2(x’ + 3) + 4(x’ + 3) (y’ + 4) + 5(y’ + 4)2<\/sup> = o
\nSimplifying this equation, we get
\n2x’2<\/sup> + 4x’y’ + 5y’2<\/sup> + 28x’ + 52y’ + 146 = 0
\nThis equation can be written (dropping dashes) as
\n2x2<\/sup> + 4xy + 5y2<\/sup> + 28x + 52y + 146 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Transformation of Axes Important Questions which are most likely to be asked in the exam. Intermediate 1st Year Maths 1B Transformation of Axes Important Questions Question 1. When the origin is shifted to (2, 3) by the translation of axes, the coordinates of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35141"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35141"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35141\/revisions"}],"predecessor-version":[{"id":35142,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35141\/revisions\/35142"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35141"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35141"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35141"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}