document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=824164>Question 824164</A>: <I><SPAN STYLE=\"word-break: break-all;\"> \"Find an equation of the locus of all points such that the difference of their distances from (6,0) and (-6,0) is always equal to 2.\" </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #496136 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=496136\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> \"Find an equation of the locus of all points such that the difference of their distances from (6,0) and (-6,0) is always equal to 2.\" tells us:<ul><li>...that the locus is a hyperbola</li><li>...that the two points from whom the difference in the distances is 2 are the foci of the hyperbola</li><li>...that the hyperbola has a horizontal transverse axis since the foci are on the same horizontal line (y = 0)</li><li>...that the center is (0, 0) since the center is always halfway between the two foci</li><li>...that the value for \"c\" for this hyperbola is 6 since the distance from the center to a focus is \"c\"</li><li>...that the value for \"a\" is 1 since the constant difference, 2, is always equal to 2a</li></ul>From all this we are almost ready to write the equation. The general equation for a horizontal hyperbola is:<BR>\n" );
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;\" ><BR>\n" );
document.write( "where \"h\" and \"k\" are the coordinates of the center. Since we already have the center and the \"a\", all we need is the \"b\".<br><BR>\n" );
document.write( "To find \"b\" we use the fact the for all hyperbolas<BR>\n" );
document.write( "<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;\" ><BR>\n" );
document.write( "Substituting in the know values for \"a\" and \"c\":<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%286%29%5E2+=+%281%29%5E2+%2B+b%5E2 ALIGN=MIDDLE  ALT=\"%286%29%5E2+=+%281%29%5E2+%2B+b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "And we solve for b (or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2 ALIGN=MIDDLE  ALT=\"b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > since that is what we need for the equation. Simplifying...<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=36+=+1+%2B+b%5E2 ALIGN=MIDDLE  ALT=\"36+=+1+%2B+b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Subtracting 1:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=35+=+b%5E2 ALIGN=MIDDLE  ALT=\"35+=+b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "We're now ready for the equation. Substituting in the values for h, k, a and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2 ALIGN=MIDDLE  ALT=\"b%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > into<BR>\n" );
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;\" ><BR>\n" );
document.write( "we get:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-%280%29%29%5E2%2F%281%29%5E2+-+%28y-%280%29%29%5E2%2F%2835%29+=+1 ALIGN=MIDDLE  ALT=\"%28x-%280%29%29%5E2%2F%281%29%5E2+-+%28y-%280%29%29%5E2%2F%2835%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Simplifying...<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2F%281%29%5E2+-+y%5E2%2F%2835%29+=+1 ALIGN=MIDDLE  ALT=\"x%5E2%2F%281%29%5E2+-+y%5E2%2F%2835%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2F1+-+y%5E2%2F%2835%29+=+1 ALIGN=MIDDLE  ALT=\"x%5E2%2F1+-+y%5E2%2F%2835%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "This is the desired equation.<br><BR>\n" );
document.write( "P.S. Although the first term simplifies down to just <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2 ALIGN=MIDDLE  ALT=\"x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > we usually leave the terms in fraction form so that the equation more closely resembles the general form. </SPAN>\n" );
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