{"id":2306,"date":"2025-08-12T21:38:49","date_gmt":"2025-08-12T21:38:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2306"},"modified":"2025-10-21T17:30:03","modified_gmt":"2025-10-21T17:30:03","slug":"parametric-equations-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/parametric-equations-learn-it-4\/","title":{"raw":"Parametric Equations: Learn It 4","rendered":"Parametric Equations: Learn It 4"},"content":{"raw":"

Eliminating the Parameter from Trigonometric Equations<\/h2>\r\nEliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.\r\n\r\nFirst, we use the identities:\r\n

[latex]\\begin{gathered}x\\left(t\\right)=a\\cos t\\\\ y\\left(t\\right)=b\\sin t\\end{gathered}[\/latex]<\/p>\r\nSolving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have\r\n

[latex]\\begin{gathered}\\frac{x}{a}=\\cos t\\\\ \\frac{y}{b}=\\sin t\\end{gathered}[\/latex]<\/p>\r\nThen, use the Pythagorean Theorem:\r\n

[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/p>\r\nSubstituting gives\r\n

[latex]{\\cos }^{2}t+{\\sin }^{2}t={\\left(\\frac{x}{a}\\right)}^{2}+{\\left(\\frac{y}{b}\\right)}^{2}=1[\/latex]<\/div>\r\n
Eliminate the parameter from the given pair of trigonometric equations<\/strong> where [latex]0\\le t\\le 2\\pi [\/latex] and sketch the graph.\r\n

[latex]\\begin{align}&x\\left(t\\right)=4\\cos t\\\\ &y\\left(t\\right)=3\\sin t\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"244567\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244567\"]\r\n\r\nSolving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have\r\n

[latex]\\begin{gathered}x=4\\cos t \\\\ \\frac{x}{4}=\\cos t \\\\ y=3\\sin t \\\\ \\frac{y}{3}=\\sin t \\end{gathered}[\/latex]<\/p>\r\nNext, use the Pythagorean identity and make the substitutions.\r\n

[latex]\\begin{gathered} {\\cos }^{2}t+{\\sin }^{2}t=1\\\\ {\\left(\\frac{x}{4}\\right)}^{2}+{\\left(\\frac{y}{3}\\right)}^{2}=1\\\\ \\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1\\end{gathered}[\/latex]<\/p>\r\nThe graph for the equation is shown.\r\n\r\n\"Graph\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\nApplying the general equations for conic sections<\/strong>, we can identify [latex]\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1[\/latex] as an ellipse centered at [latex]\\left(0,0\\right)[\/latex]. Notice that when [latex]t=0[\/latex] the coordinates are [latex]\\left(4,0\\right)[\/latex], and when [latex]t=\\frac{\\pi }{2}[\/latex] the coordinates are [latex]\\left(0,3\\right)[\/latex]. This shows the orientation of the curve with increasing values of [latex]t[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation:\r\n

[latex]x\\left(t\\right)=2\\cos t[\/latex] and [latex]y\\left(t\\right)=3\\sin t[\/latex].<\/p>\r\n[reveal-answer q=\"577648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"577648\"]\r\n\r\n[latex]\\frac{{x}^{2}}{4}+\\frac{{y}^{2}}{9}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]66927[\/ohm_question]<\/section><\/div>\r\nWhen we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially \"eliminating the parameter.\" However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as [latex]x\\left(t\\right)=t[\/latex]. In this case, [latex]y\\left(t\\right)[\/latex] can be any expression. For example, consider the following pair of equations.\r\n
[latex]\\begin{align}&x\\left(t\\right)=t\\\\ &y\\left(t\\right)={t}^{2}-3\\end{align}[\/latex]<\/div>\r\nRewriting this set of parametric equations is a matter of substituting [latex]x[\/latex] for [latex]t[\/latex]. Thus, the Cartesian equation is [latex]y={x}^{2}-3[\/latex].\r\n\r\n
Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.\r\n

[latex]\\begin{align}&x\\left(t\\right)=3t - 2 \\\\ &y\\left(t\\right)=t+1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"456091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"456091\"]\r\n\r\nMethod 1<\/em>. First, let\u2019s solve the [latex]x[\/latex] equation for [latex]t[\/latex]. Then we can substitute the result into the [latex]y[\/latex] equation.\r\n

[latex]\\begin{gathered}x=3t - 2 \\\\ x+2=3t \\\\ \\frac{x+2}{3}=t \\end{gathered}[\/latex]<\/p>\r\nNow substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.\r\n

[latex]\\begin{align}&y=t+1 \\\\ &y=\\left(\\frac{x+2}{3}\\right)+1 \\\\ &y=\\frac{x}{3}+\\frac{2}{3}+1\\\\ &y=\\frac{1}{3}x+\\frac{5}{3}\\end{align}[\/latex]<\/p>\r\nMethod 2<\/em>. Solve the [latex]y[\/latex] equation for [latex]t[\/latex] and substitute this expression in the [latex]x[\/latex] equation.\r\n

[latex]\\begin{gathered}y=t+1\\hfill \\\\ y - 1=t \\end{gathered}[\/latex]<\/p>\r\nMake the substitution and then solve for [latex]y[\/latex].\r\n

[latex]\\begin{gathered}x=3\\left(y - 1\\right)-2 \\\\ x=3y - 3-2 \\\\ x=3y - 5 \\\\ x+5=3y \\\\ \\frac{x+5}{3}=y \\\\ y=\\frac{1}{3}x+\\frac{5}{3} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Write the given parametric equations as a Cartesian equation: [latex]x\\left(t\\right)={t}^{3}[\/latex] and [latex]y\\left(t\\right)={t}^{6}[\/latex].[reveal-answer q=\"974749\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"974749\"][latex]y={x}^{2}[\/latex][\/hidden-answer]<\/section>","rendered":"

Eliminating the Parameter from Trigonometric Equations<\/h2>\n

Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.<\/p>\n

First, we use the identities:<\/p>\n

[latex]\\begin{gathered}x\\left(t\\right)=a\\cos t\\\\ y\\left(t\\right)=b\\sin t\\end{gathered}[\/latex]<\/p>\n

Solving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have<\/p>\n

[latex]\\begin{gathered}\\frac{x}{a}=\\cos t\\\\ \\frac{y}{b}=\\sin t\\end{gathered}[\/latex]<\/p>\n

Then, use the Pythagorean Theorem:<\/p>\n

[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/p>\n

Substituting gives<\/p>\n

[latex]{\\cos }^{2}t+{\\sin }^{2}t={\\left(\\frac{x}{a}\\right)}^{2}+{\\left(\\frac{y}{b}\\right)}^{2}=1[\/latex]<\/div>\n
\n
Eliminate the parameter from the given pair of trigonometric equations<\/strong> where [latex]0\\le t\\le 2\\pi[\/latex] and sketch the graph.<\/p>\n

[latex]\\begin{align}&x\\left(t\\right)=4\\cos t\\\\ &y\\left(t\\right)=3\\sin t\\end{align}[\/latex]<\/p>\n