document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=170281>Question 170281</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find Standard Equation Of a Hyperbola: Given, Foci: (-8,0) and (8,0) Vertices: (-6,0) and (6,0).  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #125693 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=125693\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Take note that ALL of the points given to you (both vertices and foci) all have a y-coordinate of 0. So this tells us that the hyperbola opens left and right like this:\r<BR>\n" );
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document.write( "Take note that the distance from the center to either focus is 8 units. So let's call this distance \"c\" (ie <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c=8 ALIGN=MIDDLE  ALT=\"c=8\"   DISABLED_style= \"max-width:90%; height:auto;\" >)\r<BR>\n" );
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document.write( "Remember, the equation of any hyperbola opening left/right is \r<BR>\n" );
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document.write( "So we need to find the values of h, k, a, and b\r<BR>\n" );
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document.write( "Now let's find the midpoint of the line connecting the vertices. This midpoint is the center of the hyperbola\r<BR>\n" );
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document.write( "x mid: Average the x-coordinates of the vertices: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-8%2B8%29%2F2+=+0%2F2+=+0 ALIGN=MIDDLE  ALT=\"%28-8%2B8%29%2F2+=+0%2F2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So the x-coordinate of the center is 0. This means that h = 0\r<BR>\n" );
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document.write( "y mid: Average the y-coordinates of the vertices: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%280%2B0%29%2F2+=+0%2F2+=+0 ALIGN=MIDDLE  ALT=\"%280%2B0%29%2F2+=+0%2F2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "So the y-coordinate of the center is 0. This means that k = 0\r<BR>\n" );
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document.write( "So the center is (0,0) which means that h=0 and k=0\r<BR>\n" );
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document.write( "Now because the hyperbola opens left and right, this means that the vertices are (h+a,k) and (h-a,k). In other words, you add and subtract the value of \"a\" to the x-coordinate of the center to get the vertices.\r<BR>\n" );
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document.write( "Since the value of \"h\" and \"k\" is 0, this means that the vertices become (0+a,0) and (0-a,0) then simplify to (a,0) and (-a,0)\r<BR>\n" );
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document.write( "So this tells us that a=6 and -a=-6 which simply means that a=6\r<BR>\n" );
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document.write( "Now it turns out that the value of \"b\" is closely connected to the values of \"a\" and \"c\". They are connected by the equation\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2%2Bb%5E2=c%5E2 ALIGN=MIDDLE  ALT=\"a%5E2%2Bb%5E2=c%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6%5E2%2Bb%5E2=8%5E2 ALIGN=MIDDLE  ALT=\"6%5E2%2Bb%5E2=8%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > Plug in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=6 ALIGN=MIDDLE  ALT=\"a=6\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c=8 ALIGN=MIDDLE  ALT=\"c=8\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=36%2Bb%5E2=64 ALIGN=MIDDLE  ALT=\"36%2Bb%5E2=64\"   DISABLED_style= \"max-width:90%; height:auto;\" > Square 6 to get 36. Square 8 to get 64\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2=64-36 ALIGN=MIDDLE  ALT=\"b%5E2=64-36\"   DISABLED_style= \"max-width:90%; height:auto;\" > Subtract 36 from both sides.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2=28 ALIGN=MIDDLE  ALT=\"b%5E2=28\"   DISABLED_style= \"max-width:90%; height:auto;\" > Subtract\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b=sqrt%2828%29 ALIGN=MIDDLE  ALT=\"b=sqrt%2828%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Take the square root of both sides. Note: only the positive square root is considered (since a negative distance doesn't make sense)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b=2%2Asqrt%287%29 ALIGN=MIDDLE  ALT=\"b=2%2Asqrt%287%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Simplify the square root.\r<BR>\n" );
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document.write( "Recap:\r<BR>\n" );
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document.write( "So we found the following: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=0 ALIGN=MIDDLE  ALT=\"h=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=k=0 ALIGN=MIDDLE  ALT=\"k=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (the x and y coordinates of the center), <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=6 ALIGN=MIDDLE  ALT=\"a=6\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b=2%2Asqrt%287%29 ALIGN=MIDDLE  ALT=\"b=2%2Asqrt%287%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28x-h%29%5E2%29%2F%28a%5E2%29-%28%28y-k%29%5E2%29%2F%28b%5E2%29=1 ALIGN=MIDDLE  ALT=\"%28%28x-h%29%5E2%29%2F%28a%5E2%29-%28%28y-k%29%5E2%29%2F%28b%5E2%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > Start with the general equation for a hyperbola (one that opens left/right)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28x-0%29%5E2%29%2F%286%5E2%29-%28%28y-0%29%5E2%29%2F%28%282%2Asqrt%287%29%29%5E2%29=1 ALIGN=MIDDLE  ALT=\"%28%28x-0%29%5E2%29%2F%286%5E2%29-%28%28y-0%29%5E2%29%2F%28%282%2Asqrt%287%29%29%5E2%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > Plug in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=0 ALIGN=MIDDLE  ALT=\"h=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=k=0 ALIGN=MIDDLE  ALT=\"k=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a=6 ALIGN=MIDDLE  ALT=\"a=6\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b=2%2Asqrt%287%29 ALIGN=MIDDLE  ALT=\"b=2%2Asqrt%287%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28x-0%29%5E2%29%2F%2836%29-%28%28y-0%29%5E2%29%2F%284%2A7%29=1 ALIGN=MIDDLE  ALT=\"%28%28x-0%29%5E2%29%2F%2836%29-%28%28y-0%29%5E2%29%2F%284%2A7%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > Square 6 to get 36. Square <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2Asqrt%287%29 ALIGN=MIDDLE  ALT=\"2%2Asqrt%287%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4%2A7 ALIGN=MIDDLE  ALT=\"4%2A7\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28x-0%29%5E2%29%2F%2836%29-%28%28y-0%29%5E2%29%2F%2828%29=1 ALIGN=MIDDLE  ALT=\"%28%28x-0%29%5E2%29%2F%2836%29-%28%28y-0%29%5E2%29%2F%2828%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > Multiply\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2%29%2F%2836%29-%28y%5E2%29%2F%2828%29=1 ALIGN=MIDDLE  ALT=\"%28x%5E2%29%2F%2836%29-%28y%5E2%29%2F%2828%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" > Simplify\r<BR>\n" );
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document.write( "Answer:\r<BR>\n" );
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document.write( "So the equation of the hyperbola that has the foci (8,0) and (-8,0) along with the vertices (-6,0) and (6,0) is:\r<BR>\n" );
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