document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=298194>Question 298194</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Write the equation. \r<BR>\n" );
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document.write( "Hyperbola with C(-2,0)  V(-2,3)  y= +/- .5x\r<BR>\n" );
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document.write( "I am pretty sure C stands for center and V for vertex. There are two vertices so the second one should be at (-2,-3) I think. I am not sure what to do with the y = +/- .5x though.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #214623 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=214623\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The general equation for a horizontally oriented hyperbola would be:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-h%29%5E2%2Fa%5E2+=+%28y-k%29%5E2%2Fb%5E2+=+1 ALIGN=MIDDLE  ALT=\"%28x-h%29%5E2%2Fa%5E2+=+%28y-k%29%5E2%2Fb%5E2+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The center of your hyperbola is at C = (-2,0).\r<BR>\n" );
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document.write( "The vertices of this hyperbola are probably at V1 = (-2,-3) and V2 = (-2,3)\r<BR>\n" );
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document.write( "This, I believe, makes this a vertically oriented hyperbola whose general equation would be:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1 ALIGN=MIDDLE  ALT=\"%28y-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The y = +/- .5x refers to the equation of the asymptotes for the hyperbola.\r<BR>\n" );
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document.write( "In a horizontally oriented hyperbola, that equation would be:\r<BR>\n" );
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document.write( "y = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28b%2Fa%29%2A%28x-h%29+%2B+k ALIGN=MIDDLE  ALT=\"%28b%2Fa%29%2A%28x-h%29+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "In a vertically oriented hyperbola, that equation would be:\r<BR>\n" );
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document.write( "y = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%2Fb%29%2A%28x-h%29+%2B+k ALIGN=MIDDLE  ALT=\"%28a%2Fb%29%2A%28x-h%29+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The distance between the center of the hyperbola and each vertex of the hyperbola is equal to \"a\".\r<BR>\n" );
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document.write( "Since your center is (-2,0) and one of your vertices is at (2,3), then \"a\" must be equal to 3.\r<BR>\n" );
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document.write( "Since the general equation of the asymptotes of a vertically oriented hyperbola is:\r<BR>\n" );
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document.write( "y = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%2Fb%29+%2A+%28x-h%29+%2B+k ALIGN=MIDDLE  ALT=\"%28a%2Fb%29+%2A+%28x-h%29+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" > and the equation of the asymptotes of your equation is y = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=.5+%2A+x ALIGN=MIDDLE  ALT=\".5+%2A+x\"   DISABLED_style= \"max-width:90%; height:auto;\" >, then the equation of the asymptotes for your hyperbola becomes:\r<BR>\n" );
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document.write( "y = +/- .5x = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%2Fb%29+%2A+%28x-h%29+%2B+k ALIGN=MIDDLE  ALT=\"%28a%2Fb%29+%2A+%28x-h%29+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "This makes:\r<BR>\n" );
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document.write( "+/- .5x = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%2Fb%29+%2A+%28x-h%29+%2B+k ALIGN=MIDDLE  ALT=\"%28a%2Fb%29+%2A+%28x-h%29+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The slope of the asymptote on the left side of this equation is +/- .5\r<BR>\n" );
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document.write( "The slope of the asymptote on the right side of this equation is +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2Fb ALIGN=MIDDLE  ALT=\"a%2Fb\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Since a = +/- 3, this makes the slope of the asymptote on the right side of this equation equal to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2Fb%29 ALIGN=MIDDLE  ALT=\"3%2Fb%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Because these slopes are just different versions of the slope of the same asymptotes, we can solve for b as follows:\r<BR>\n" );
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document.write( "+/- .5 = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2Fb ALIGN=MIDDLE  ALT=\"3%2Fb\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Multiply both sides of this equation by b to get:\r<BR>\n" );
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document.write( "+/- .5b = +/- 3\r<BR>\n" );
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document.write( "Multiply both sides of this equation by 2 to get:\r<BR>\n" );
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document.write( "+/- b = +/- 6\r<BR>\n" );
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document.write( "We now have:\r<BR>\n" );
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document.write( "a = 3<BR>\n" );
document.write( "b = 6\r<BR>\n" );
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document.write( "Since the general form of the equation of our vertically oriented hyperbola is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1 ALIGN=MIDDLE  ALT=\"%28y-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we can replace a^2 with 9 and b^2 with 36 to get:\r<BR>\n" );
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document.write( "Since the center of our hyperbola is at (-2,0), and the center of the hyperbola is represented by (h,k), this means that:\r<BR>\n" );
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document.write( "h = -2<BR>\n" );
document.write( "k = 0\r<BR>\n" );
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document.write( "We can replace h and k in the equation of our hyperbola to get:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2%2F9+-+%28x%2B2%29%5E2%2F36+=+1 ALIGN=MIDDLE  ALT=\"y%5E2%2F9+-+%28x%2B2%29%5E2%2F36+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "That's the equation of our hyperbola.\r<BR>\n" );
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document.write( "To graph this equation, we have to solve for y.\r<BR>\n" );
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document.write( "Solving for y gets us:\r<BR>\n" );
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document.write( "y = +/- <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29 ALIGN=MIDDLE  ALT=\"sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The graph of our equation looks like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-9%2C9%2C-9%2C9%2Csqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C-sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "A more distant view looks like this:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-36%2C36%2C-36%2C36%2Csqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C-sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The equation for the asymptotes of this hyperbola are y = +.- .5\r<BR>\n" );
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document.write( "We add the equations of those lines to the distant graph to get:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-36%2C36%2C-36%2C36%2Csqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C-sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C.5x%2C-.5x%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The asymptotes are a little off because we did not account for the center of the hyperbola.\r<BR>\n" );
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document.write( "Apparently, the equation y = +/- .5x was the equation for the slope of the asymptotes only.\r<BR>\n" );
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document.write( "Since the asymptotes have to go through the center of the hyperbola, we need to modify this equation to account for the fact that the lines are going through the point (-2,0).\r<BR>\n" );
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document.write( "The point slope form of the equation of a straight line is (y-y1) = m*(x-x1)\r<BR>\n" );
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document.write( "The slope of our line is +/- .5\r<BR>\n" );
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document.write( "The point slope form of the equation of the asymptotes of our hyperbola becomes:\r<BR>\n" );
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document.write( "(y-y1) = +/- .5*(x-x1)\r<BR>\n" );
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document.write( "Since the lines goes through the point (-2,0), then (x1,y1) = (-2,0) and the equation of our asymptotes becomes:\r<BR>\n" );
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document.write( "y = +/- .5*(x+2)\r<BR>\n" );
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document.write( "We graph these equations of our asymptotes on top of the equation of our hyperbola to get:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-36%2C36%2C-36%2C36%2Csqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C-sqrt%289%2A%281%2B%28%28x%2B2%29%5E2%2F36%29%29%29%2C.5x%2B1%2C-.5x-1%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now everything lines up right and the graph clearly shows the asymptotes of the equation of our hyperbola.\r<BR>\n" );
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document.write( "You did not need to graph the asymptotes.   That's a bonus.\r<BR>\n" );
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document.write( "Another bonus is the eccentricity of the hyperbola.\r<BR>\n" );
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document.write( "It is given by the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=e+=+c%2Fa ALIGN=MIDDLE  ALT=\"e+=+c%2Fa\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "c is the measure of the distance from the center of the hyperbola to one of the foci of the hyperbola.\r<BR>\n" );
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document.write( "The value of c is given by the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c%5E2+=+a%5E2+%2B+b%5E2 ALIGN=MIDDLE  ALT=\"c%5E2+=+a%5E2+%2B+b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "In this hyperbola, that would make <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c+=+sqrt%289%2B36%29+=+sqrt%2845%29 ALIGN=MIDDLE  ALT=\"c+=+sqrt%289%2B36%29+=+sqrt%2845%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The eccentricity of this hyperbola is therefore <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=e+=+c%2Fa+=+sqrt%2845%29%2F3 ALIGN=MIDDLE  ALT=\"e+=+c%2Fa+=+sqrt%2845%29%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 2.236067978\r<BR>\n" );
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document.write( "The higher the eccentricity, the straighter is the hyperbola.   A very very high eccentricity would make the hyperbola look more like two parallel lines.\r<BR>\n" );
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document.write( "This one is somewhere in the medium range.\r<BR>\n" );
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document.write( "\"a\" and \"c\" are defined as:\r<BR>\n" );
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document.write( "a is the distance from the center of the hyperbola to each vertex.<BR>\n" );
document.write( "c is the distance from the center of the hyperbola to each focus.\r<BR>\n" );
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document.write( "A pretty decent reference for you to look at is shown below:\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.purplemath.com/modules/hyperbola.htm\" target = \"_blank\">http://www.purplemath.com/modules/hyperbola.htm</a>\r<BR>\n" );
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