document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168978>Question 1168978</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Foci at (-1,4) and (7,4), transverse axis has the length 8/3. Whats the equation  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851463 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851463\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Solution:<BR>\n" );
document.write( "The given information indicates that the conic section is a hyperbola because we are given foci and a transverse axis length.\r<BR>\n" );
document.write( "\n" );
document.write( "The foci are at $F_1(-1, 4)$ and $F_2(7, 4)$.<BR>\n" );
document.write( "The transverse axis has a length of $\frac{8}{3}$.\r<BR>\n" );
document.write( "\n" );
document.write( "From the foci, we can determine the center of the hyperbola as the midpoint of the foci:<BR>\n" );
document.write( "Center $(h, k) = \left( \frac{-1 + 7}{2}, \frac{4 + 4}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)$.\r<BR>\n" );
document.write( "\n" );
document.write( "The distance between the foci is $2c$.<BR>\n" );
document.write( "$2c = \sqrt{(7 - (-1))^2 + (4 - 4)^2} = \sqrt{(8)^2 + (0)^2} = \sqrt{64} = 8$.<BR>\n" );
document.write( "So, $c = 4$.\r<BR>\n" );
document.write( "\n" );
document.write( "The length of the transverse axis is $2a$. We are given $2a = \frac{8}{3}$, so $a = \frac{4}{3}$.\r<BR>\n" );
document.write( "\n" );
document.write( "Since the y-coordinates of the foci are the same, the transverse axis is horizontal. The standard form of the equation of a hyperbola with a horizontal transverse axis and center $(h, k)$ is:<BR>\n" );
document.write( "$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$\r<BR>\n" );
document.write( "\n" );
document.write( "We need to find $b^2$. We use the relationship $c^2 = a^2 + b^2$ for a hyperbola.<BR>\n" );
document.write( "$4^2 = \left(\frac{4}{3}\right)^2 + b^2$<BR>\n" );
document.write( "$16 = \frac{16}{9} + b^2$<BR>\n" );
document.write( "$b^2 = 16 - \frac{16}{9} = \frac{144 - 16}{9} = \frac{128}{9}$.\r<BR>\n" );
document.write( "\n" );
document.write( "Now, substitute the values of $h=3$, $k=4$, $a^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$, and $b^2 = \frac{128}{9}$ into the standard form:<BR>\n" );
document.write( "$\frac{(x - 3)^2}{\frac{16}{9}} - \frac{(y - 4)^2}{\frac{128}{9}} = 1$\r<BR>\n" );
document.write( "\n" );
document.write( "Multiply the numerators by $\frac{9}{16}$ and $\frac{9}{128}$ respectively:<BR>\n" );
document.write( "$\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1$.\r<BR>\n" );
document.write( "\n" );
document.write( "Final Answer: The final answer is $\boxed{\frac{9(x - 3)^2}{16} - \frac{9(y - 4)^2}{128} = 1}$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );