{"id":2570,"date":"2025-08-13T18:07:27","date_gmt":"2025-08-13T18:07:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2570"},"modified":"2025-08-13T18:08:25","modified_gmt":"2025-08-13T18:08:25","slug":"rotation-of-axes-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/rotation-of-axes-learn-it-3\/","title":{"raw":"Rotation of Axes: Learn It 3","rendered":"Rotation of Axes: Learn It 3"},"content":{"raw":"

Writing Equations of Rotated Conics in Standard Form<\/h2>\r\nNow that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex] into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] coordinate system without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term, by rotating the axes by a measure of [latex]\\theta [\/latex] that satisfies\r\n
[latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex]<\/div>\r\nWe have learned already that any conic may be represented by the second degree equation\r\n
[latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex]<\/div>\r\nwhere [latex]A,B[\/latex], and [latex]C[\/latex] are not all zero. However, if [latex]B\\ne 0[\/latex], then we have an [latex]xy[\/latex] term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle [latex]\\theta [\/latex] where [latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex].\r\n
\r\n
    \r\n \t
  • If [latex]\\cot \\left(2\\theta \\right)>0[\/latex], then [latex]2\\theta [\/latex] is in the first quadrant, and [latex]\\theta [\/latex] is between [latex]\\left(0^\\circ ,45^\\circ \\right)[\/latex].<\/li>\r\n \t
  • If [latex]\\cot \\left(2\\theta \\right)<0[\/latex], then [latex]2\\theta [\/latex] is in the second quadrant, and [latex]\\theta [\/latex] is between [latex]\\left(45^\\circ ,90^\\circ \\right)[\/latex].<\/li>\r\n \t
  • If [latex]A=C[\/latex], then [latex]\\theta =45^\\circ [\/latex].<\/li>\r\n<\/ul>\r\n
    How To: Given an equation for a conic in the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system, rewrite the equation without the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] term in terms of [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex], where the [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex] axes are rotations of the standard axes by [latex]\\theta [\/latex] degrees.<\/strong>\r\n
      \r\n \t
    1. Find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/li>\r\n \t
    2. Find [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex].<\/li>\r\n \t
    3. Substitute [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex] into [latex]\\begin{align}x={x}^{\\prime }\\cos \\theta -{y}^{\\prime }\\sin \\theta \\end{align}[\/latex] and [latex]\\begin{align} y={x}^{\\prime }\\sin \\theta +{y}^{\\prime }\\cos \\theta \\end{align}[\/latex].<\/li>\r\n \t
    4. Substitute the expression for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.<\/li>\r\n \t
    5. Write the equations with [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex] in the standard form with respect to the rotated axes.<\/li>\r\n<\/ol>\r\n<\/section>
      Rewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[\/latex] in the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system without an [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] term.[reveal-answer q=\"952851\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"952851\"]First, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].\r\n

      [latex]\\begin{align}8{x}^{2}-12xy+17{y}^{2}=20\\Rightarrow A=8,B=-12\\text{ and }C=17\\end{align}[\/latex]<\/p>\r\n

      [latex]\\begin{align} &\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}=\\frac{8 - 17}{-12} \\\\ &\\cot \\left(2\\theta \\right)=\\frac{-9}{-12}=\\frac{3}{4} \\end{align}[\/latex]<\/p>\r\n\"\"\r\n

      <\/div>\r\n

      [latex]\\cot \\left(2\\theta \\right)=\\frac{3}{4}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/p>\r\nSo the hypotenuse is\r\n

      [latex]\\begin{gathered} {3}^{2}+{4}^{2}={h}^{2}\\\\ 9+16={h}^{2}\\\\ 25={h}^{2}\\\\ h=5\\end{gathered}[\/latex]<\/p>\r\nNext, we find [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex].\r\n

      [latex]\\begin{align} \\sin \\theta &=\\sqrt{\\frac{1-\\cos \\left(2\\theta \\right)}{2}} \\\\ &=\\sqrt{\\frac{1-\\frac{3}{5}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{5}{5}-\\frac{3}{5}}{2}} \\\\ &=\\sqrt{\\frac{5 - 3}{5}\\cdot \\frac{1}{2}} \\\\ &=\\sqrt{\\frac{2}{10}} \\\\ &=\\sqrt{\\frac{1}{5}} \\\\ &=\\frac{1}{\\sqrt{5}} \\\\[2mm] \\cos \\theta &=\\sqrt{\\frac{1+\\cos \\left(2\\theta \\right)}{2}} \\\\ &=\\sqrt{\\frac{1+\\frac{3}{5}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{5}{5}+\\frac{3}{5}}{2}} \\\\ &=\\sqrt{\\frac{5+3}{5}\\cdot \\frac{1}{2}} \\\\ &=\\sqrt{\\frac{8}{10}} \\\\ &=\\sqrt{\\frac{4}{5}} \\\\ &=\\frac{2}{\\sqrt{5}} \\end{align}[\/latex]<\/p>\r\nSubstitute the values of [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex] into [latex]\\begin{align}x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta \\end{align}[\/latex] and [latex]\\begin{align}y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta \\end{align}[\/latex].\r\n

      [latex]\\begin{align}&x={x}^{\\prime }\\cos \\theta -{y}^{\\prime }\\sin \\theta \\\\ &x={x}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right)-{y}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right) \\\\ &x=\\frac{2{x}^{\\prime }-{y}^{\\prime }}{\\sqrt{5}} \\end{align}[\/latex]<\/p>\r\nand\r\n

      [latex]\\begin{align} &y={x}^{\\prime }\\sin \\theta +{y}^{\\prime }\\cos \\theta \\\\ &y={x}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right)+{y}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right) \\\\ &y=\\frac{{x}^{\\prime }+2{y}^{\\prime }}{\\sqrt{5}} \\end{align}[\/latex]<\/p>\r\nSubstitute the expressions for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.\r\n

      [latex]\\begin{gathered}8{\\left(\\frac{2{x}^{\\prime }-{y}^{\\prime }}{\\sqrt{5}}\\right)}^{2}-12\\left(\\frac{2{x}^{\\prime }-{y}^{\\prime }}{\\sqrt{5}}\\right)\\left(\\frac{{x}^{\\prime }+2{y}^{\\prime }}{\\sqrt{5}}\\right)+17{\\left(\\frac{{x}^{\\prime }+2{y}^{\\prime }}{\\sqrt{5}}\\right)}^{2}=20 \\\\ 8\\left(\\frac{\\left(2{x}^{\\prime }-{y}^{\\prime }\\right)\\left(2{x}^{\\prime }-{y}^{\\prime }\\right)}{5}\\right)-12\\left(\\frac{\\left(2{x}^{\\prime }-{y}^{\\prime }\\right)\\left({x}^{\\prime }+2{y}^{\\prime }\\right)}{5}\\right)+17\\left(\\frac{\\left({x}^{\\prime }+2{y}^{\\prime }\\right)\\left({x}^{\\prime }+2{y}^{\\prime }\\right)}{5}\\right)=20 \\\\ 8\\left(4{x}^{\\prime }{}^{2}-4{x}^{\\prime }{y}^{\\prime }+{y}^{\\prime }{}^{2}\\right)-12\\left(2{x}^{\\prime }{}^{2}+3{x}^{\\prime }{y}^{\\prime }-2{y}^{\\prime }{}^{2}\\right)+17\\left({x}^{\\prime }{}^{2}+4{x}^{\\prime }{y}^{\\prime }+4{y}^{\\prime }{}^{2}\\right)=100 \\\\ 32{x}^{\\prime }{}^{2}-32{x}^{\\prime }{y}^{\\prime }+8{y}^{\\prime }{}^{2}-24{x}^{\\prime }{}^{2}-36{x}^{\\prime }{y}^{\\prime }+24{y}^{\\prime }{}^{2}+17{x}^{\\prime }{}^{2}+68{x}^{\\prime }{y}^{\\prime }+68{y}^{\\prime }{}^{2}=100 \\\\ 25{x}^{\\prime }{}^{2}+100{y}^{\\prime }{}^{2}=100 \\\\ \\frac{25}{100}{x}^{\\prime }{}^{2}+\\frac{100}{100}{y}^{\\prime }{}^{2}=\\frac{100}{100} \\end{gathered}[\/latex]<\/p>\r\nWrite the equations with [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex] in the standard form with respect to the new coordinate system.\r\n

      [latex]\\begin{align}\\frac{{{x}^{\\prime }}^{2}}{4}+\\frac{{{y}^{\\prime }}^{2}}{1}=1\\end{align}[\/latex]<\/p>\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      Rewrite the [latex]13{x}^{2}-6\\sqrt{3}xy+7{y}^{2}=16[\/latex] in the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system without the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] term.[reveal-answer q=\"452974\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"452974\"][latex]\\begin{align}\\frac{{{x}^{\\prime }}^{2}}{4}+\\frac{{{y}^{\\prime }}^{2}}{1}=1\\end{align}[\/latex][\/hidden-answer]<\/section>
      Graph the following equation relative to the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system:\r\n

      [latex]{x}^{2}+12xy - 4{y}^{2}=30[\/latex]<\/p>\r\n[reveal-answer q=\"16063\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16063\"]\r\n\r\nFirst, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].\r\n

      [latex]{x}^{2}+12xy - 4{y}^{2}=20\\Rightarrow A=1,B=12,\\text{ and }C=-4[\/latex]<\/p>\r\n

      [latex]\\begin{align}&\\cot \\left(2\\theta \\right)=\\frac{A-C}{B} \\\\ &\\cot \\left(2\\theta \\right)=\\frac{1-\\left(-4\\right)}{12} \\\\ &\\cot \\left(2\\theta \\right)=\\frac{5}{12} \\end{align}[\/latex]<\/p>\r\nBecause [latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}[\/latex], we can draw a reference triangle.\r\n\r\n\"\"\r\n

      <\/div>\r\n

      [latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/p>\r\nThus, the hypotenuse is\r\n

      [latex]\\begin{gathered} {5}^{2}+{12}^{2}={h}^{2}\\\\ 25+144={h}^{2}\\\\ 169={h}^{2}\\\\ h=13\\end{gathered}[\/latex]<\/p>\r\nNext, we find [latex]\\sin \\theta [\/latex] and [latex]\\cos \\theta [\/latex]. We will use half-angle identities.\r\n

      [latex]\\begin{align} \\sin \\theta &=\\sqrt{\\frac{1-\\cos \\left(2\\theta \\right)}{2}} \\\\ &=\\sqrt{\\frac{1-\\frac{5}{13}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{13}{13}-\\frac{5}{13}}{2}} \\\\&=\\sqrt{\\frac{8}{13}\\cdot \\frac{1}{2}} \\\\ &=\\frac{2}{\\sqrt{13}} \\\\ \\cos \\theta &=\\sqrt{\\frac{1+\\cos \\left(2\\theta \\right)}{2}} \\\\ &=\\sqrt{\\frac{1+\\frac{5}{13}}{2}} \\\\ &=\\sqrt{\\frac{\\frac{13}{13}+\\frac{5}{13}}{2}} \\\\ &=\\sqrt{\\frac{18}{13}\\cdot \\frac{1}{2}} \\\\ &=\\frac{3}{\\sqrt{13}} \\end{align}[\/latex]<\/p>\r\nNow we find [latex]x[\/latex] and [latex]y[\/latex].\r\n

      [latex]\\begin{align} &x={x}^{\\prime }\\cos \\theta -{y}^{\\prime }\\sin \\theta \\\\ &x={x}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right)-{y}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right) \\\\ &x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}} \\end{align}[\/latex]<\/p>\r\nand\r\n

      [latex]\\begin{align} &y={x}^{\\prime }\\sin \\theta +{y}^{\\prime }\\cos \\theta \\\\ &y={x}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right)+{y}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right) \\\\ &y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}} \\end{align}[\/latex]<\/p>\r\nNow we substitute [latex]\\begin{align}x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}\\end{align}[\/latex] and [latex]\\begin{align}y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}\\end{align}[\/latex] into [latex]\\begin{align}{x}^{2}+12xy - 4{y}^{2}=30\\end{align}[\/latex].\r\n

      [latex]\\begin{align} &{\\left(\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}\\right)}^{2}+12\\left(\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}\\right)\\left(\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}\\right)-4{\\left(\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}\\right)}^{2}=30 \\\\ &\\left(\\frac{1}{13}\\right)\\left[{\\left(3{x}^{\\prime }-2{y}^{\\prime }\\right)}^{2}+12\\left(3{x}^{\\prime }-2{y}^{\\prime }\\right)\\left(2{x}^{\\prime }+3{y}^{\\prime }\\right)-4{\\left(2{x}^{\\prime }+3{y}^{\\prime }\\right)}^{2}\\right]=30 && \\text{Factor}. \\\\ &\\left(\\frac{1}{13}\\right)\\left[9{x}^{\\prime }{}^{2}-12{x}^{\\prime }{y}^{\\prime }+4{y}^{\\prime }{}^{2}+12\\left(6{x}^{\\prime }{}^{2}+5{x}^{\\prime }{y}^{\\prime }-6{y}^{\\prime }{}^{2}\\right)-4\\left(4{x}^{\\prime }{}^{2}+12{x}^{\\prime }{y}^{\\prime }+9{y}^{\\prime }{}^{2}\\right)\\right]=30 && \\text{Multiply}. \\\\ &\\left(\\frac{1}{13}\\right)\\left[9{x}^{\\prime }{}^{2}-12{x}^{\\prime }{y}^{\\prime }+4{y}^{\\prime }{}^{2}+72{x}^{\\prime }{}^{2}+60{x}^{\\prime }{y}^{\\prime }-72{y}^{\\prime }{}^{2}-16{x}^{\\prime }{}^{2}-48{x}^{\\prime }{y}^{\\prime }-36{y}^{\\prime }{}^{2}\\right]=30 && \\text{Distribute}. \\\\ &\\left(\\frac{1}{13}\\right)\\left[65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}\\right]=30 && \\text{Combine like terms}. \\\\ &65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}=390 && \\text{Multiply}. \\\\ &\\frac{{x}^{\\prime }{}^{2}}{6}-\\frac{4{y}^{\\prime }{}^{2}}{15}=1 && \\text{Divide by 390}. \\end{align}[\/latex]<\/p>\r\nThe graph of the hyperbola [latex]\\begin{align}\\frac{{{x}^{\\prime }}^{2}}{6}-\\frac{4{{y}^{\\prime }}^{2}}{15}=1\\end{align}[\/latex].\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

      Writing Equations of Rotated Conics in Standard Form<\/h2>\n

      Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex] into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] coordinate system without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term, by rotating the axes by a measure of [latex]\\theta[\/latex] that satisfies<\/p>\n

      [latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex]<\/div>\n

      We have learned already that any conic may be represented by the second degree equation<\/p>\n

      [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex]<\/div>\n

      where [latex]A,B[\/latex], and [latex]C[\/latex] are not all zero. However, if [latex]B\\ne 0[\/latex], then we have an [latex]xy[\/latex] term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle [latex]\\theta[\/latex] where [latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex].<\/p>\n

      \n
        \n
      • If [latex]\\cot \\left(2\\theta \\right)>0[\/latex], then [latex]2\\theta[\/latex] is in the first quadrant, and [latex]\\theta[\/latex] is between [latex]\\left(0^\\circ ,45^\\circ \\right)[\/latex].<\/li>\n
      • If [latex]\\cot \\left(2\\theta \\right)<0[\/latex], then [latex]2\\theta[\/latex] is in the second quadrant, and [latex]\\theta[\/latex] is between [latex]\\left(45^\\circ ,90^\\circ \\right)[\/latex].<\/li>\n
      • If [latex]A=C[\/latex], then [latex]\\theta =45^\\circ[\/latex].<\/li>\n<\/ul>\n
        How To: Given an equation for a conic in the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system, rewrite the equation without the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] term in terms of [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex], where the [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex] axes are rotations of the standard axes by [latex]\\theta[\/latex] degrees.<\/strong><\/p>\n
          \n
        1. Find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/li>\n
        2. Find [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex].<\/li>\n
        3. Substitute [latex]\\sin \\theta[\/latex] and [latex]\\cos \\theta[\/latex] into [latex]\\begin{align}x={x}^{\\prime }\\cos \\theta -{y}^{\\prime }\\sin \\theta \\end{align}[\/latex] and [latex]\\begin{align} y={x}^{\\prime }\\sin \\theta +{y}^{\\prime }\\cos \\theta \\end{align}[\/latex].<\/li>\n
        4. Substitute the expression for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.<\/li>\n
        5. Write the equations with [latex]\\begin{align}{x}^{\\prime }\\end{align}[\/latex] and [latex]\\begin{align}{y}^{\\prime }\\end{align}[\/latex] in the standard form with respect to the rotated axes.<\/li>\n<\/ol>\n<\/section>\n
          Rewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[\/latex] in the [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] system without an [latex]\\begin{align}{x}^{\\prime }{y}^{\\prime }\\end{align}[\/latex] term.<\/p>\n