document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=975398>Question 975398</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For the following hyperbola:  \r<BR>\n" );
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document.write( "            (x + 1)2  –  (y – 3)2 = 1                                   <BR>\n" );
document.write( "                16              9\r<BR>\n" );
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document.write( "Find: a)Center :___________\r<BR>\n" );
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document.write( "         b)Vertices:___________<BR>\n" );
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document.write( "                           ___________\r<BR>\n" );
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document.write( "         c)  Foci:     ___________\r<BR>\n" );
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document.write( "                                                     ___________<BR>\n" );
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document.write( "        d) Asymptotes: (list separately)\r<BR>\n" );
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document.write( "            _________________\r<BR>\n" );
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document.write( "            _________________      <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #597161 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=597161\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For the following hyperbola: <BR>\n" );
document.write( "(x+1)^2/16 – (y–3)^2/9 = 1 <BR>\n" );
document.write( "Find: a)Center : (-1, 3)<BR>\n" );
document.write( "b)Vertices:  (-3, 3) and (5, 3)<BR>\n" );
document.write( "c) Foci: (-4, 3) and (6, 3)<BR>\n" );
document.write( "d) Asymptotes: (list separately)<BR>\n" );
document.write( " y=-3x/4+9/4 <BR>\n" );
document.write( " y=3x/4+15/4<BR>\n" );
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document.write( "Given hyperbola has a horizontal transverse axis.<BR>\n" );
document.write( "Its standard form of equation: <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;\" >, (h,k)=coordinates of center.<BR>\n" );
document.write( "..<BR>\n" );
document.write( "center: (-1, 3)<BR>\n" );
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document.write( "a^2=16<BR>\n" );
document.write( "a=&#8730;16=4<BR>\n" );
document.write( "vertices: (-1±a, 3)=(1±4, 3)= (-3, 3) and (5, 3)<BR>\n" );
document.write( "..<BR>\n" );
document.write( "b^2=9<BR>\n" );
document.write( "b=&#8730;9=3<BR>\n" );
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document.write( "c^2=a^2+b^2=16+9=25<BR>\n" );
document.write( "c=&#8730;25=5<BR>\n" );
document.write( "..<BR>\n" );
document.write( "foci: (-1±c, 3)= (1±5, 3= (-4, 3) and (6, 3)<BR>\n" );
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document.write( "Asymptotes are straight lines that go thru center.<BR>\n" );
document.write( "slopes of asymptotes=±b/a=±3/4<BR>\n" );
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document.write( "Equation of asymptote with negative slope:<BR>\n" );
document.write( "y=-3x/4+b<BR>\n" );
document.write( "solve for b using coordinates of center (-1, 3) which are on the line<BR>\n" );
document.write( "b=y+3x/4=3+(3*(-1)/4)=9/4<BR>\n" );
document.write( "b=9/4<BR>\n" );
document.write( "equation: y=-3x/4+9/4<BR>\n" );
document.write( "..<BR>\n" );
document.write( "Equation of asymptote with positive slope:<BR>\n" );
document.write( "y=3x/4+b<BR>\n" );
document.write( "solve for b using coordinates of center (-1, 3) which are on the line<BR>\n" );
document.write( "b=y-3x/4=3-(3*(-1)/4)=9/4<BR>\n" );
document.write( "b=15/4<BR>\n" );
document.write( "equation:  y=3x/4+15/4\r<BR>\n" );
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