document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1121384>Question 1121384</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Convert the equation to standard form. Locate the foci and find the equation of the asymptotes.<BR>\n" );
document.write( "16x2 + 128x - 9y2 + 180y - 788 = 0  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #737290 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=737290\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "(1) Move the constant to the other side of the equation; factor out the leading coefficients in x and y:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16%28x%5E2%2B8x%29-9%28y%5E2-20y%29+=+788 ALIGN=MIDDLE  ALT=\"16%28x%5E2%2B8x%29-9%28y%5E2-20y%29+=+788\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "(2) Complete the square in both x and y, remembering to add the same amounts to both sides of the equation:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16%28x%5E2%2B8x%2B16%29-9%28y%5E2-20y%2B100%29+=+788%2B256-900+=+144 ALIGN=MIDDLE  ALT=\"16%28x%5E2%2B8x%2B16%29-9%28y%5E2-20y%2B100%29+=+788%2B256-900+=+144\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "(3) Write the trinomials as binomials squared; divide through by 144 to get \"1\" on the right side:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+1 ALIGN=MIDDLE  ALT=\"%28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "or<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B4%29%5E2%2F3%5E2-%28y-10%29%5E2%2F4%5E2+=+1 ALIGN=MIDDLE  ALT=\"%28x%2B4%29%5E2%2F3%5E2-%28y-10%29%5E2%2F4%5E2+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "This is standard form,<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><BR>\n" );
document.write( "The center is (h,k) = (-4,10); a=3 is the distance from the center to each vertex; b=4 is the distance from the center to each co-vertex.<br><BR>\n" );
document.write( "The distance from the center to each focus is c, where for a hyperbola <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=c%5E2+=+a%5E2%2Bb%5E2 ALIGN=MIDDLE  ALT=\"c%5E2+=+a%5E2%2Bb%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  So c=5.<br><BR>\n" );
document.write( "Answers:<BR>\n" );
document.write( "(a) The equation is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+1 ALIGN=MIDDLE  ALT=\"%28x%2B4%29%5E2%2F9-%28y-10%29%5E2%2F16+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "(b) The branches of the hyperbola open left and right; The foci are a distance c=5 right or left of the center, at (-9,10) and (1,10).<BR>\n" );
document.write( "(c) The slopes of the asymptotes are b/a and -b/a; in this example, 4/3 and -4/3.  Use the point-slope form of the equation of a line using each of those slopes, using the center (-4,10) as the point.<BR>\n" );
document.write( " </SPAN>\n" );
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