document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=470703>Question 470703</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The question I have for homework is\r<BR>\n" );
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document.write( "Find an equation of the hyperbola having foci at (-1,1-sqrt52), and -1,1+sqrt52) and asymptotes at  y=2/3x+5/3 and y=-2/3x+1/3\r<BR>\n" );
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document.write( "I know c = sqrt of 52 and c^2 = 52 I am just unsure on how to find A and B. I know how to graph the asymptotes and I know the center is (-1,1)\r<BR>\n" );
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document.write( "I know I have to use 2/3 for A and B I just dont understand how. Any help would be greatly appreciated thanks.. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #323439 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=323439\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find an equation of the hyperbola having foci at (-1,1-sqrt52), and -1,1+sqrt52) and asymptotes at y=2/3x+5/3 and y=-2/3x+1/3 <BR>\n" );
document.write( "I know c = sqrt of 52 and c^2 = 52 I am just unsure on how to find A and B. I know how to graph the asymptotes and I know the center is (-1,1) <BR>\n" );
document.write( "I know I have to use 2/3 for A and B I just dont understand how.<BR>\n" );
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document.write( "From given Foci coordinates you can see that this is a hyperbola with vertical transverse axis. <BR>\n" );
document.write( "(y changes but x does not)<BR>\n" );
document.write( "Standard form for hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being (x,y) coordinates of center<BR>\n" );
document.write( "Slope, m, for asymptotes of this form=a/b<BR>\n" );
document.write( "From given asymptote equations, m=±2/3<BR>\n" );
document.write( "a/b=2/3<BR>\n" );
document.write( "b=3a/2<BR>\n" );
document.write( "b^2=9a^2/4<BR>\n" );
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document.write( "From Foci<BR>\n" );
document.write( "c=&#8730;52<BR>\n" );
document.write( "c^2=a^2+b^2<BR>\n" );
document.write( "52=a^2+9a^2/4<BR>\n" );
document.write( "208=4a^2+9a^2=/13a^2<BR>\n" );
document.write( "a^2=208/13=16<BR>\n" );
document.write( "b^2=9a^2/4=9*16/4=36<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "(y-1)^2/16-(x+1)^2/36=1<BR>\n" );
document.write( "Asymptotes:<BR>\n" );
document.write( "2x/3+5/3<BR>\n" );
document.write( "-2x/3+1/3<BR>\n" );
document.write( "see graph below as a visual check on the answer<BR>\n" );
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document.write( "y=±(16+16(x+1)^2/36)^.5+1<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C%2816%2B16%28x%2B1%29%5E2%2F36%29%5E.5%2B1%2C-%2816%2B16%28x%2B1%29%5E2%2F36%29%5E.5%2B1%2C%0D%0A-2x%2F3%2B1%2F3%2C2x%2F3%2B5%2F3%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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