{"id":2601,"date":"2024-08-08T01:15:50","date_gmt":"2024-08-08T01:15:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2601"},"modified":"2024-11-21T22:37:56","modified_gmt":"2024-11-21T22:37:56","slug":"hyperbola-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/hyperbola-learn-it-3\/","title":{"raw":"Hyperbola: Learn It 3","rendered":"Hyperbola: Learn It 3"},"content":{"raw":"

Hyperbolas Not Centered at the Origin<\/h2>\r\n
\r\n
\r\n
\r\n\r\nLike the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated units horizontally and units vertically, the center of the hyperbola will be [latex](h,k)[\/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex](x-h)[\/latex] and [latex]y[\/latex] replaced by [latex](y-k)[\/latex].<\/span>\r\n\r\n
\r\n

standard forms of the equation of a hyperbola with center [latex](h,k)[\/latex]<\/h3>\r\nThe standard form of the equation of a hyperbola with center [latex]\\left(h,k\\right)[\/latex] and transverse axis parallel to the [latex]x[\/latex]-axis is\r\n\r\n
[latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/center>\r\nwhere\r\n
    \r\n \t
  • the length of the transverse axis is [latex]2a[\/latex]<\/li>\r\n \t
  • the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\r\n \t
  • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\r\n \t
  • the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\r\n \t
  • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\r\n \t
  • the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\r\n<\/ul>\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"\" (a) Horizontal hyperbola with center [latex]\\left(h,k\\right)[\/latex] (b) Vertical hyperbola with center [latex]\\left(h,k\\right)[\/latex][\/caption]<\/section>Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.\r\n\r\n
    How To: Given the vertices and foci of a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], write its equation in standard form.<\/strong>\r\n
      \r\n \t
    1. Determine whether the transverse axis is parallel to the [latex]x[\/latex]- or [latex]y[\/latex]-axis.\r\n
        \r\n \t
      1. If the [latex]y[\/latex]-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the [latex]x[\/latex]-axis. Use the standard form [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n \t
      2. If the [latex]x[\/latex]-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the [latex]y[\/latex]-axis. Use the standard form [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. Identify the center of the hyperbola, [latex]\\left(h,k\\right)[\/latex], using the midpoint formula and the given coordinates for the vertices.<\/li>\r\n \t
      4. Find [latex]{a}^{2}[\/latex] by solving for the length of the transverse axis, [latex]2a[\/latex] , which is the distance between the given vertices.<\/li>\r\n \t
      5. Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex] found in Step 2 along with the given coordinates for the foci.<\/li>\r\n \t
      6. Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex].<\/li>\r\n \t
      7. Substitute the values for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\r\n<\/ol>\r\n<\/section>
        What is the standard form equation of the hyperbola<\/strong> that has vertices at [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex] and foci at [latex]\\left(-2,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)?[\/latex][reveal-answer q=\"360650\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"360650\"]The y<\/em>-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x<\/em>-axis. Thus, the equation of the hyperbola will have the form\r\n\r\n \r\n

        [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/p>\r\nFirst, we identify the center, [latex]\\left(h,k\\right)[\/latex]. The center is halfway between the vertices [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex]. Applying the midpoint formula, we have\r\n

        [latex]\\left(h,k\\right)=\\left(\\dfrac{0+6}{2},\\dfrac{-2+\\left(-2\\right)}{2}\\right)=\\left(3,-2\\right)[\/latex]<\/p>\r\nNext, we find [latex]{a}^{2}[\/latex]. The length of the transverse axis, [latex]2a[\/latex], is bounded by the vertices. So, we can find [latex]{a}^{2}[\/latex] by finding the distance between the x<\/em>-coordinates of the vertices.\r\n

        [latex]\\begin{gathered}2a=|0 - 6| \\\\ 2a=6 \\\\ a=3 \\\\ {a}^{2}=9 \\end{gathered}[\/latex]<\/p>\r\nNow we need to find [latex]{c}^{2}[\/latex]. The coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]. So [latex]\\left(h-c,k\\right)=\\left(-2,-2\\right)[\/latex] and [latex]\\left(h+c,k\\right)=\\left(8,-2\\right)[\/latex]. We can use the x<\/em>-coordinate from either of these points to solve for [latex]c[\/latex]. Using the point [latex]\\left(8,-2\\right)[\/latex], and substituting [latex]h=3[\/latex],\r\n

        [latex]\\begin{gathered}h+c=8 \\\\ 3+c=8 \\\\ c=5 \\\\ {c}^{2}=25 \\end{gathered}[\/latex]<\/p>\r\nNext, solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}:[\/latex]\r\n

        [latex]\\begin{align}{b}^{2}&={c}^{2}-{a}^{2} \\\\ &=25 - 9 \\\\ &=16 \\end{align}[\/latex]<\/p>\r\nFinally, substitute the values found for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation.\r\n

        [latex]\\dfrac{{\\left(x - 3\\right)}^{2}}{9}-\\dfrac{{\\left(y+2\\right)}^{2}}{16}=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>","rendered":"

        Hyperbolas Not Centered at the Origin<\/h2>\n
        \n
        \n
        \n

        Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated units horizontally and units vertically, the center of the hyperbola will be [latex](h,k)[\/latex]. This translation results in the standard form of the equation we saw previously, with [latex]x[\/latex] replaced by [latex](x-h)[\/latex] and [latex]y[\/latex] replaced by [latex](y-k)[\/latex].<\/span><\/p>\n

        \n

        standard forms of the equation of a hyperbola with center [latex](h,k)[\/latex]<\/h3>\n

        The standard form of the equation of a hyperbola with center [latex]\\left(h,k\\right)[\/latex] and transverse axis parallel to the [latex]x[\/latex]-axis is<\/p>\n

        [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex]<\/div>\n

        where<\/p>\n

          \n
        • the length of the transverse axis is [latex]2a[\/latex]<\/li>\n
        • the coordinates of the vertices are [latex]\\left(h\\pm a,k\\right)[\/latex]<\/li>\n
        • the length of the conjugate axis is [latex]2b[\/latex]<\/li>\n
        • the coordinates of the co-vertices are [latex]\\left(h,k\\pm b\\right)[\/latex]<\/li>\n
        • the distance between the foci is [latex]2c[\/latex], where [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]<\/li>\n
        • the coordinates of the foci are [latex]\\left(h\\pm c,k\\right)[\/latex]<\/li>\n<\/ul>\n
          \"\"
          (a) Horizontal hyperbola with center [latex]\\left(h,k\\right)[\/latex] (b) Vertical hyperbola with center [latex]\\left(h,k\\right)[\/latex]<\/figcaption><\/figure>\n<\/section>\n

          Like hyperbolas centered at the origin, hyperbolas centered at a point [latex]\\left(h,k\\right)[\/latex] have vertices, co-vertices, and foci that are related by the equation [latex]{c}^{2}={a}^{2}+{b}^{2}[\/latex]. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given.<\/p>\n

          How To: Given the vertices and foci of a hyperbola centered at [latex]\\left(h,k\\right)[\/latex], write its equation in standard form.<\/strong><\/p>\n
            \n
          1. Determine whether the transverse axis is parallel to the [latex]x[\/latex]– or [latex]y[\/latex]-axis.\n
              \n
            1. If the [latex]y[\/latex]-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the [latex]x[\/latex]-axis. Use the standard form [latex]\\dfrac{{\\left(x-h\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(y-k\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n
            2. If the [latex]x[\/latex]-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the [latex]y[\/latex]-axis. Use the standard form [latex]\\dfrac{{\\left(y-k\\right)}^{2}}{{a}^{2}}-\\dfrac{{\\left(x-h\\right)}^{2}}{{b}^{2}}=1[\/latex].<\/li>\n<\/ol>\n<\/li>\n
            3. Identify the center of the hyperbola, [latex]\\left(h,k\\right)[\/latex], using the midpoint formula and the given coordinates for the vertices.<\/li>\n
            4. Find [latex]{a}^{2}[\/latex] by solving for the length of the transverse axis, [latex]2a[\/latex] , which is the distance between the given vertices.<\/li>\n
            5. Find [latex]{c}^{2}[\/latex] using [latex]h[\/latex] and [latex]k[\/latex] found in Step 2 along with the given coordinates for the foci.<\/li>\n
            6. Solve for [latex]{b}^{2}[\/latex] using the equation [latex]{b}^{2}={c}^{2}-{a}^{2}[\/latex].<\/li>\n
            7. Substitute the values for [latex]h,k,{a}^{2}[\/latex], and [latex]{b}^{2}[\/latex] into the standard form of the equation determined in Step 1.<\/li>\n<\/ol>\n<\/section>\n
              What is the standard form equation of the hyperbola<\/strong> that has vertices at [latex]\\left(0,-2\\right)[\/latex] and [latex]\\left(6,-2\\right)[\/latex] and foci at [latex]\\left(-2,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)?[\/latex]<\/p>\n