document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=449662>Question 449662</A>: <I><SPAN STYLE=\"word-break: break-all;\"> write the equation of the conic section with the given information.\r<BR>\n" );
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document.write( "hyperbola with vertices (3,-2) and (-9,-2), and foci (7, -2) and (-13, -2) <BR>\n" );
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document.write( "Thank you so much for all your help!!! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #309792 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=309792\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> write the equation of the conic section with the given information. <BR>\n" );
document.write( "hyperbola with vertices (3,-2) and (-9,-2), and foci (7, -2) and (-13, -2) <BR>\n" );
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document.write( "Standard form for hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1,<BR>\n" );
document.write( "with (h,k) being the (x,y) coordinates of the center.<BR>\n" );
document.write( "Standard form for hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1<BR>\n" );
document.write( "with (h,k) being the (x,y) coordinates of the center.<BR>\n" );
document.write( "Note that the only difference between these two forms is that the x and y terms are interchanged.<BR>\n" );
document.write( "..<BR>\n" );
document.write( "The best way to start solving conics problems is drawing a sketch of the conic with the given information like the four points of the vertices and foci.<BR>\n" );
document.write( "Since the vertices and foci are on the line y=-2, transverse axis of given hyperbola is horizontal, that is, it opens left-right. (First form listed above)<BR>\n" );
document.write( "Center of hyperbola is also on the line y=-2, so its y-coordinate=-2.<BR>\n" );
document.write( "The x-coordinate of center is the mid point of the end points of the vertices=(3+(-9))/2=6/2=-3.<BR>\n" );
document.write( "So, coordinates of center is at (-3,-2)<BR>\n" );
document.write( "Length of vertex is the distance between the end points of a vertex=12=2a<BR>\n" );
document.write( "a=6<BR>\n" );
document.write( "a^2=36<BR>\n" );
document.write( "c=distance from center to one of the foci=10<BR>\n" );
document.write( "c^2=100<BR>\n" );
document.write( "c^2=a^2+b^2<BR>\n" );
document.write( "b^2=c^2-a^2=100-36=64<BR>\n" );
document.write( "b=8<BR>\n" );
document.write( "We now have the information we need to write the equation of given hyperbola in standard form<BR>\n" );
document.write( " (x-h)^2/a^2-(y-k)^2/b^2=1<BR>\n" );
document.write( " (x+3)^2/36-(y+2)^2/64=1  (ans)<BR>\n" );
document.write( "See the graph of given hyperbola below which should help you relate the visual evidence to the algebra above.\r<BR>\n" );
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document.write( "..<BR>\n" );
document.write( "(64((x+3)^2/36-1))^.5-2<BR>\n" );
document.write( "(8/6)(x+3)-2 (equation of asymptotes)\r<BR>\n" );
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