{"id":1871,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1871"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"deriving-the-equation-of-a-hyperbola-centered-at-the-origin","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/deriving-the-equation-of-a-hyperbola-centered-at-the-origin\/","title":{"raw":"Deriving the Equation of a Hyperbola Centered at the Origin","rendered":"Deriving the Equation of a Hyperbola Centered at the Origin"},"content":{"raw":"

Let [latex]\\left(-c,0\\right)[\/latex] and [latex]\\left(c,0\\right)[\/latex] be the foci<\/strong> of a hyperbola centered at the origin. The hyperbola is the set of all points [latex]\\left(x,y\\right)[\/latex] such that the difference of the distances from [latex]\\left(x,y\\right)[\/latex] to the foci is constant.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 4<\/b>[\/caption]\n\nIf [latex]\\left(a,0\\right)[\/latex] is a vertex of the hyperbola, the distance from [latex]\\left(-c,0\\right)[\/latex] to [latex]\\left(a,0\\right)[\/latex] is [latex]a-\\left(-c\\right)=a+c[\/latex]. The distance from [latex]\\left(c,0\\right)[\/latex] to [latex]\\left(a,0\\right)[\/latex] is [latex]c-a[\/latex]. The sum of the distances from the foci to the vertex is\n<\/p>

[latex]\\left(a+c\\right)-\\left(c-a\\right)=2a[\/latex]<\/div>\nIf [latex]\\left(x,y\\right)[\/latex] is a point on the hyperbola, we can define the following variables:\n
[latex]\\begin{array}{l}{d}_{2}=\\text{the distance from }\\left(-c,0\\right)\\text{ to }\\left(x,y\\right)\\\\ {d}_{1}=\\text{the distance from }\\left(c,0\\right)\\text{ to }\\left(x,y\\right)\\end{array}[\/latex]<\/div>\nBy definition of a hyperbola, [latex]{d}_{2}-{d}_{1}[\/latex] is constant for any point [latex]\\left(x,y\\right)[\/latex] on the hyperbola. We know that the difference of these distances is [latex]2a[\/latex] for the vertex [latex]\\left(a,0\\right)[\/latex]. It follows that [latex]{d}_{2}-{d}_{1}=2a[\/latex] for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula<\/strong>. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses.\n
[latex]\\begin{array}{ll}\\text{ }{d}_{2}-{d}_{1}=\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y - 0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y - 0\\right)}^{2}}=2a\\hfill & \\text{Distance Formula}\\hfill \\\\ \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}=2a\\hfill & \\text{Simplify expressions}\\text{.}\\hfill \\\\ \\text{ }\\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}=2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill & \\text{Move radical to opposite side}\\text{.}\\hfill \\\\ \\text{ }{\\left(x+c\\right)}^{2}+{y}^{2}={\\left(2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right)}^{2}\\hfill & \\text{Square both sides}\\text{.}\\hfill \\\\ \\text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{\\left(x-c\\right)}^{2}+{y}^{2}\\hfill & \\text{Expand the squares}\\text{.}\\hfill \\\\ \\text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{c}^{2}+{y}^{2}\\hfill & \\text{Expand remaining square}\\text{.}\\hfill \\\\ \\text{ }2cx=4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}-2cx\\hfill & \\text{Combine like terms}\\text{.}\\hfill \\\\ \\text{ }4cx - 4{a}^{2}=4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill & \\text{Isolate the radical}\\text{.}\\hfill \\\\ \\text{ }cx-{a}^{2}=a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill & \\text{Divide by 4}\\text{.}\\hfill \\\\ \\text{ }{\\left(cx-{a}^{2}\\right)}^{2}={a}^{2}{\\left[\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right]}^{2}\\hfill & \\text{Square both sides}\\text{.}\\hfill \\\\ \\text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}\\left({x}^{2}-2cx+{c}^{2}+{y}^{2}\\right)\\hfill & \\text{Expand the squares}\\text{.}\\hfill \\\\ \\text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}{x}^{2}-2{a}^{2}cx+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill & \\text{Distribute }{a}^{2}\\text{.}\\hfill \\\\ \\text{ }{a}^{4}+{c}^{2}{x}^{2}={a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill & \\text{Combine like terms}\\text{.}\\hfill \\\\ \\text{ }{c}^{2}{x}^{2}-{a}^{2}{x}^{2}-{a}^{2}{y}^{2}={a}^{2}{c}^{2}-{a}^{4}\\hfill & \\text{Rearrange terms}\\text{.}\\hfill \\\\ \\text{ }{x}^{2}\\left({c}^{2}-{a}^{2}\\right)-{a}^{2}{y}^{2}={a}^{2}\\left({c}^{2}-{a}^{2}\\right)\\hfill & \\text{Factor common terms}\\text{.}\\hfill \\\\ \\text{ }{x}^{2}{b}^{2}-{a}^{2}{y}^{2}={a}^{2}{b}^{2}\\hfill & \\text{Set }{b}^{2}={c}^{2}-{a}^{2}.\\hfill \\\\ \\text{ }\\frac{{x}^{2}{b}^{2}}{{a}^{2}{b}^{2}}-\\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}=\\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}\\hfill & \\text{Divide both sides by }{a}^{2}{b}^{2}\\hfill \\\\ \\text{ }\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1\\hfill & \\hfill \\end{array}[\/latex]<\/div>\nThis equation defines a hyperbola centered at the origin with vertices [latex]\\left(\\pm a,0\\right)[\/latex] and co-vertices [latex]\\left(0\\pm b\\right)[\/latex].\n
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A General Note: Standard Forms of the Equation of a Hyperbola with Center (0,0)<\/h3>\nThe standard form of the equation of a hyperbola with center [latex]\\left(0,0\\right)[\/latex] and transverse axis on the x<\/em>-axis is\n
[latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex]<\/div>\nwhere\n