{"id":1960,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1960"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"writing-equations-of-rotated-conics-in-standard-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/writing-equations-of-rotated-conics-in-standard-form\/","title":{"raw":"Writing Equations of Rotated Conics in Standard Form","rendered":"Writing Equations of Rotated Conics in Standard Form"},"content":{"raw":"<p>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\n<\/p><div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex]<\/div>\nWe have learned already that any conic may be represented by the second degree equation\n<div style=\"text-align: center;\">[latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex]<\/div>\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].\n<div>\n<ul><li>If [latex]\\cot \\left(2\\theta \\right)&gt;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\t<li>If [latex]\\cot \\left(2\\theta \\right)&lt;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\t<li>If [latex]A=C[\/latex], then [latex]\\theta =45^\\circ [\/latex].<\/li>\n<\/ul><\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation for a conic in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system, rewrite the equation without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term in terms of [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex], where the [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] axes are rotations of the standard axes by [latex]\\theta [\/latex] degrees.<\/h3>\n<ol><li>Find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/li>\n\t<li>Find [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex].<\/li>\n\t<li>Substitute [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex] into [latex]x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta [\/latex] and [latex]y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta [\/latex].<\/li>\n\t<li>Substitute the expression for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.<\/li>\n\t<li>Write the equations with [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] in the standard form with respect to the rotated axes.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Rewriting an Equation with respect to the <em>x\u2032<\/em> and <em>y\u2032<\/em> axes without the <em>x\u2032y\u2032<\/em> Term<\/h3>\nRewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[\/latex] in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system without an [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nFirst, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}8{x}^{2}-12xy+17{y}^{2}=20\\Rightarrow A=8,B=-12\\text{and}C=17\\hfill \\\\ \\text{ }\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}=\\frac{8 - 17}{-12}\\hfill \\\\ \\text{ }\\cot \\left(2\\theta \\right)=\\frac{-9}{-12}=\\frac{3}{4}\\hfill \\end{array}[\/latex]<\/div>\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202410\/CNX_Precalc_Figure_10_04_0072.jpg\" alt=\"\" width=\"487\" height=\"328\" data-media-type=\"image\/jpg\"\/><b>Figure 7<\/b>[\/caption]\n\n<div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{3}{4}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/div>\nSo the hypotenuse is\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill {3}^{2}+{4}^{2}={h}^{2}\\\\ \\hfill 9+16={h}^{2}\\\\ \\hfill 25={h}^{2}\\\\ \\hfill h=5\\end{array}[\/latex]<\/div>\nNext, we find [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ \\sin \\text{ }\\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}}\\hfill \\end{array}\\hfill \\\\ \\sin \\text{ }\\theta =\\frac{1}{\\sqrt{5}}\\hfill \\\\ \\cos \\text{ }\\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}}\\hfill \\\\ \\cos \\text{ }\\theta =\\frac{2}{\\sqrt{5}}\\hfill \\end{array}[\/latex]<\/div>\nSubstitute the values of [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex] into [latex]x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta [\/latex] and [latex]y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta [\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta \\hfill \\\\ x={x}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right)-{y}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right)\\hfill \\\\ x=\\frac{2{x}^{\\prime }-{y}^{\\prime }}{\\sqrt{5}}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\nand\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta \\hfill \\end{array}\\hfill \\\\ y={x}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right)+{y}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right)\\hfill \\\\ y=\\frac{{x}^{\\prime }+2{y}^{\\prime }}{\\sqrt{5}}\\hfill \\end{array}[\/latex]<\/div>\nSubstitute the expressions for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }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\\text{ }\\hfill \\\\ \\text{ }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\\text{ }\\hfill \\\\ \\text{ }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\\hfill \\\\ 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\\hfill \\\\ \\text{ }25{x}^{\\prime }{}^{2}+100{y}^{\\prime }{}^{2}=100\\text{ }\\hfill \\\\ \\text{ }\\frac{25}{100}{x}^{\\prime }{}^{2}+\\frac{100}{100}{y}^{\\prime }{}^{2}=\\frac{100}{100} \\hfill \\end{array}[\/latex]<\/div>\nWrite the equations with [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] in the standard form with respect to the new coordinate system.\n<div style=\"text-align: center;\">[latex]\\frac{{{x}^{\\prime }}^{2}}{4}+\\frac{{{y}^{\\prime }}^{2}}{1}=1[\/latex]<\/div>\nFigure 8 shows the graph of the ellipse.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202411\/CNX_Precalc_Figure_10_04_0082.jpg\" alt=\"\" width=\"487\" height=\"217\" data-media-type=\"image\/jpg\"\/><b>Figure 8<\/b>[\/caption]\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\nRewrite the [latex]13{x}^{2}-6\\sqrt{3}xy+7{y}^{2}=16[\/latex] in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term.\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-26\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Graphing an Equation That Has No <em>x\u2032y\u2032<\/em> Terms<\/h3>\nGraph the following equation relative to the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system:\n<div style=\"text-align: center;\">[latex]{x}^{2}+12xy - 4{y}^{2}=30[\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\nFirst, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].\n<div style=\"text-align: center;\">[latex]{x}^{2}+12xy - 4{y}^{2}=20\\Rightarrow A=1,\\text{ }B=12,\\text{and }C=-4[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}\\hfill \\\\ \\cot \\left(2\\theta \\right)=\\frac{1-\\left(-4\\right)}{12}\\hfill \\\\ \\cot \\left(2\\theta \\right)=\\frac{5}{12}\\hfill \\end{array}[\/latex]<\/div>\nBecause [latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}[\/latex], we can draw a reference triangle as in Figure 9.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202412\/CNX_Precalc_Figure_10_04_0092.jpg\" alt=\"\" width=\"487\" height=\"591\" data-media-type=\"image\/jpg\"\/><b>Figure 9<\/b>[\/caption]\n\n<div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/div>\nThus, the hypotenuse is\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill {5}^{2}+{12}^{2}={h}^{2}\\\\ \\hfill 25+144={h}^{2}\\\\ \\hfill 169={h}^{2}\\\\ \\hfill h=13\\end{array}[\/latex]<\/div>\nNext, we find [latex]\\sin \\text{ }\\theta [\/latex] and [latex]\\cos \\text{ }\\theta [\/latex]. We will use half-angle identities.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ \\sin \\text{ }\\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}}\\hfill \\end{array}\\hfill \\\\ \\cos \\text{ }\\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}}\\hfill \\end{array}[\/latex]<\/div>\nNow we find [latex]x[\/latex] and [latex]y\\text{.\\hspace{0.17em}}[\/latex]\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta \\hfill \\\\ x={x}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right)-{y}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right)\\hfill \\\\ x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}\\hfill \\end{array}[\/latex]<\/div>\nand\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta \\hfill \\\\ y={x}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right)+{y}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right)\\hfill \\\\ y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}\\hfill \\end{array}[\/latex]<\/div>\nNow we substitute [latex]x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}[\/latex] and [latex]y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}[\/latex] into [latex]{x}^{2}+12xy - 4{y}^{2}=30[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }{\\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\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }\\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 \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Factor}.\\hfill \\\\ \\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\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Multiply}.\\hfill \\\\ \\text{ }\\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\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Distribute}.\\hfill \\\\ \\text{ }\\text{ }\\left(\\frac{1}{13}\\right)\\left[65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}\\right]=30\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ \\text{ }65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}=390\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Multiply}.\\text{ }\\hfill \\\\ \\text{ }\\frac{{x}^{\\prime }{}^{2}}{6}-\\frac{4{y}^{\\prime }{}^{2}}{15}=1 \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Divide by 390}.\\hfill \\end{array}[\/latex]<\/div>\nFigure 10 shows the graph of the hyperbola [latex]\\frac{{{x}^{\\prime }}^{2}}{6}-\\frac{4{{y}^{\\prime }}^{2}}{15}=1.\\text{ }[\/latex]\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202413\/CNX_Precalc_Figure_10_04_0102.jpg\" alt=\"\" width=\"487\" height=\"441\" data-media-type=\"image\/jpg\"\/><b>Figure 10<\/b>[\/caption]\n\n<\/div>","rendered":"<p>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\n<\/p>\n<div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}[\/latex]<\/div>\n<p>We have learned already that any conic may be represented by the second degree equation<\/p>\n<div style=\"text-align: center;\">[latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[\/latex]<\/div>\n<p>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<div>\n<ul>\n<li>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<li>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<li>If [latex]A=C[\/latex], then [latex]\\theta =45^\\circ[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation for a conic in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system, rewrite the equation without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term in terms of [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex], where the [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] axes are rotations of the standard axes by [latex]\\theta[\/latex] degrees.<\/h3>\n<ol>\n<li>Find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/li>\n<li>Find [latex]\\sin \\text{ }\\theta[\/latex] and [latex]\\cos \\text{ }\\theta[\/latex].<\/li>\n<li>Substitute [latex]\\sin \\text{ }\\theta[\/latex] and [latex]\\cos \\text{ }\\theta[\/latex] into [latex]x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta[\/latex] and [latex]y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta[\/latex].<\/li>\n<li>Substitute the expression for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.<\/li>\n<li>Write the equations with [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] in the standard form with respect to the rotated axes.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Rewriting an Equation with respect to the <em>x\u2032<\/em> and <em>y\u2032<\/em> axes without the <em>x\u2032y\u2032<\/em> Term<\/h3>\n<p>Rewrite the equation [latex]8{x}^{2}-12xy+17{y}^{2}=20[\/latex] in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system without an [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>First, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}8{x}^{2}-12xy+17{y}^{2}=20\\Rightarrow A=8,B=-12\\text{and}C=17\\hfill \\\\ \\text{ }\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}=\\frac{8 - 17}{-12}\\hfill \\\\ \\text{ }\\cot \\left(2\\theta \\right)=\\frac{-9}{-12}=\\frac{3}{4}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202410\/CNX_Precalc_Figure_10_04_0072.jpg\" alt=\"\" width=\"487\" height=\"328\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{3}{4}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/div>\n<p>So the hypotenuse is<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill {3}^{2}+{4}^{2}={h}^{2}\\\\ \\hfill 9+16={h}^{2}\\\\ \\hfill 25={h}^{2}\\\\ \\hfill h=5\\end{array}[\/latex]<\/div>\n<p>Next, we find [latex]\\sin \\text{ }\\theta[\/latex] and [latex]\\cos \\text{ }\\theta[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ \\sin \\text{ }\\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}}\\hfill \\end{array}\\hfill \\\\ \\sin \\text{ }\\theta =\\frac{1}{\\sqrt{5}}\\hfill \\\\ \\cos \\text{ }\\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}}\\hfill \\\\ \\cos \\text{ }\\theta =\\frac{2}{\\sqrt{5}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Substitute the values of [latex]\\sin \\text{ }\\theta[\/latex] and [latex]\\cos \\text{ }\\theta[\/latex] into [latex]x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta[\/latex] and [latex]y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta \\hfill \\\\ x={x}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right)-{y}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right)\\hfill \\\\ x=\\frac{2{x}^{\\prime }-{y}^{\\prime }}{\\sqrt{5}}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>and<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta \\hfill \\end{array}\\hfill \\\\ y={x}^{\\prime }\\left(\\frac{1}{\\sqrt{5}}\\right)+{y}^{\\prime }\\left(\\frac{2}{\\sqrt{5}}\\right)\\hfill \\\\ y=\\frac{{x}^{\\prime }+2{y}^{\\prime }}{\\sqrt{5}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Substitute the expressions for [latex]x[\/latex] and [latex]y[\/latex] into in the given equation, and then simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }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\\text{ }\\hfill \\\\ \\text{ }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\\text{ }\\hfill \\\\ \\text{ }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\\hfill \\\\ 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\\hfill \\\\ \\text{ }25{x}^{\\prime }{}^{2}+100{y}^{\\prime }{}^{2}=100\\text{ }\\hfill \\\\ \\text{ }\\frac{25}{100}{x}^{\\prime }{}^{2}+\\frac{100}{100}{y}^{\\prime }{}^{2}=\\frac{100}{100} \\hfill \\end{array}[\/latex]<\/div>\n<p>Write the equations with [latex]{x}^{\\prime }[\/latex] and [latex]{y}^{\\prime }[\/latex] in the standard form with respect to the new coordinate system.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{{{x}^{\\prime }}^{2}}{4}+\\frac{{{y}^{\\prime }}^{2}}{1}=1[\/latex]<\/div>\n<p>Figure 8 shows the graph of the ellipse.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202411\/CNX_Precalc_Figure_10_04_0082.jpg\" alt=\"\" width=\"487\" height=\"217\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p>Rewrite the [latex]13{x}^{2}-6\\sqrt{3}xy+7{y}^{2}=16[\/latex] in the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system without the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] term.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-26\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Graphing an Equation That Has No <em>x\u2032y\u2032<\/em> Terms<\/h3>\n<p>Graph the following equation relative to the [latex]{x}^{\\prime }{y}^{\\prime }[\/latex] system:<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}+12xy - 4{y}^{2}=30[\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<p>First, we find [latex]\\cot \\left(2\\theta \\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{x}^{2}+12xy - 4{y}^{2}=20\\Rightarrow A=1,\\text{ }B=12,\\text{and }C=-4[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\cot \\left(2\\theta \\right)=\\frac{A-C}{B}\\hfill \\\\ \\cot \\left(2\\theta \\right)=\\frac{1-\\left(-4\\right)}{12}\\hfill \\\\ \\cot \\left(2\\theta \\right)=\\frac{5}{12}\\hfill \\end{array}[\/latex]<\/div>\n<p>Because [latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}[\/latex], we can draw a reference triangle as in Figure 9.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202412\/CNX_Precalc_Figure_10_04_0092.jpg\" alt=\"\" width=\"487\" height=\"591\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div style=\"text-align: center;\">[latex]\\cot \\left(2\\theta \\right)=\\frac{5}{12}=\\frac{\\text{adjacent}}{\\text{opposite}}[\/latex]<\/div>\n<p>Thus, the hypotenuse is<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill {5}^{2}+{12}^{2}={h}^{2}\\\\ \\hfill 25+144={h}^{2}\\\\ \\hfill 169={h}^{2}\\\\ \\hfill h=13\\end{array}[\/latex]<\/div>\n<p>Next, we find [latex]\\sin \\text{ }\\theta[\/latex] and [latex]\\cos \\text{ }\\theta[\/latex]. We will use half-angle identities.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ \\sin \\text{ }\\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}}\\hfill \\end{array}\\hfill \\\\ \\cos \\text{ }\\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}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Now we find [latex]x[\/latex] and [latex]y\\text{.\\hspace{0.17em}}[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ x={x}^{\\prime }\\cos \\text{ }\\theta -{y}^{\\prime }\\sin \\text{ }\\theta \\hfill \\\\ x={x}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right)-{y}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right)\\hfill \\\\ x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}\\hfill \\end{array}[\/latex]<\/div>\n<p>and<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\hfill \\\\ y={x}^{\\prime }\\sin \\text{ }\\theta +{y}^{\\prime }\\cos \\text{ }\\theta \\hfill \\\\ y={x}^{\\prime }\\left(\\frac{2}{\\sqrt{13}}\\right)+{y}^{\\prime }\\left(\\frac{3}{\\sqrt{13}}\\right)\\hfill \\\\ y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Now we substitute [latex]x=\\frac{3{x}^{\\prime }-2{y}^{\\prime }}{\\sqrt{13}}[\/latex] and [latex]y=\\frac{2{x}^{\\prime }+3{y}^{\\prime }}{\\sqrt{13}}[\/latex] into [latex]{x}^{2}+12xy - 4{y}^{2}=30[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }{\\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\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }\\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 \\hfill & \\hfill & \\hfill & \\text{Factor}.\\hfill \\\\ \\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\\hfill & \\hfill & \\hfill & \\text{Multiply}.\\hfill \\\\ \\text{ }\\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\\hfill & \\hfill & \\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }\\text{ }\\left(\\frac{1}{13}\\right)\\left[65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}\\right]=30\\hfill & \\hfill & \\hfill & \\text{Combine like terms}.\\hfill \\\\ \\text{ }65{x}^{\\prime }{}^{2}-104{y}^{\\prime }{}^{2}=390\\hfill & \\hfill & \\hfill & \\text{Multiply}.\\text{ }\\hfill \\\\ \\text{ }\\frac{{x}^{\\prime }{}^{2}}{6}-\\frac{4{y}^{\\prime }{}^{2}}{15}=1 \\hfill & \\hfill & \\hfill & \\text{Divide by 390}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Figure 10 shows the graph of the hyperbola [latex]\\frac{{{x}^{\\prime }}^{2}}{6}-\\frac{4{{y}^{\\prime }}^{2}}{15}=1.\\text{ }[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202413\/CNX_Precalc_Figure_10_04_0102.jpg\" alt=\"\" width=\"487\" height=\"441\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1960\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax 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