document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=77990>Question 77990</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I can't figure out this hyperbola: (((3x^2 + y^2 +18x - 2y + 4=0))). This is what I've tried so far:  3x^2/3 + y^2 + 18x/3 - 2y +4=0<BR>\n" );
document.write( "                         x^2 +y^2 + 6x - 2y =-4<BR>\n" );
document.write( "                         x^2/-4 + y^2/-4 + -3x/2 + y/2 = 1)))<BR>\n" );
document.write( "please help me  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #55896 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=55896\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> since the squared terms are both positive, this is an ellipse rather than a hyperbola...start by regrouping terms\r<BR>\n" );
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document.write( "3(x^2+6x+___)+y^2-2y+___=-4...now complete the squares 3(x^2+6x+9)+y^2-2y+1=-4+27+1...so 3(x+3)^2+(y-1)^2=24\r<BR>\n" );
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document.write( "dividing by 24 gives <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28%28x%2B3%29%5E2%29%2F8%29%2B%28%28%28y-1%29%5E2%29%2F24%29=1 ALIGN=MIDDLE  ALT=\"%28%28%28x%2B3%29%5E2%29%2F8%29%2B%28%28%28y-1%29%5E2%29%2F24%29=1\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "an ellipse centered at (-3,1); with the foci at (-3,-3) and (-3,5) </SPAN>\n" );
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