document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=449074>Question 449074</A>: <I><SPAN STYLE=\"word-break: break-all;\"> determine an equation for the hyberbola with center (0,-2), one vertex at (0,-4) and one foci at (0,2) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #309109 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=309109\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> determine an equation for the hyberbola with center (0,-2), one vertex at (0,-4) and one foci at (0,2)<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Standard form for a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1<BR>\n" );
document.write( "Standard form for a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1<BR>\n" );
document.write( "Note that the only difference between these two forms is that (x-h)^2 and (y-k)^2 are interchanged.<BR>\n" );
document.write( "...<BR>\n" );
document.write( "Since given vertex is vertical, hyperbola is of the first form, that is, it has a vertical transverse axis.<BR>\n" );
document.write( "a=distance from center to one of the vertices on the transverse axis=2<BR>\n" );
document.write( "a^2=4<BR>\n" );
document.write( "c=distance from center to one of the foci on the transverse axis=4<BR>\n" );
document.write( "c^2=16<BR>\n" );
document.write( "c^2=a^2+b^2<BR>\n" );
document.write( "b^2=c^2-a^2=16-4=12<BR>\n" );
document.write( "b=&#8730;12=3.46.. <BR>\n" );
document.write( "you now have enough information to write the equation for given hyperbola as follows:<BR>\n" );
document.write( "ctr (0,-2)<BR>\n" );
document.write( "a^2=4<BR>\n" );
document.write( "b^2=12\r<BR>\n" );
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document.write( "(y+2)^2/4-(x-0)^2/12=1<BR>\n" );
document.write( "(y+2)^2/4-x^2/12=1<BR>\n" );
document.write( "See graph below as visual evidence of given hyperbola:<BR>\n" );
document.write( "..<BR>\n" );
document.write( "y=(4(1+x^2/12))^.5-2<BR>\n" );
document.write( "equation of asymptotes =±(4x/6.92)-2<BR>\n" );
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