document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165937>Question 1165937</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve for the equation of the hyperbola in general form with directrices x=-5 and x=7, and one focus at (13,5). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #852876 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.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=852876\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To find the equation of the hyperbola in general form, we first need to determine the **center**, the distance from the center to the **focus ($c$)**, and the distance from the center to the **directrix ($a/e$)**.\r<BR>\n" );
document.write( "\n" );
document.write( "Since the directrices are vertical lines ($x=-5$ and $x=7$), the hyperbola is **horizontal**, meaning its transverse axis is parallel to the $x$-axis.\r<BR>\n" );
document.write( "\n" );
document.write( "## 1. Determine Key Parameters\r<BR>\n" );
document.write( "\n" );
document.write( "### A. Center ($h, k$)\r<BR>\n" );
document.write( "\n" );
document.write( "The center ($h, k$) lies exactly halfway between the two directrices. Since the directrices are defined by $x=-5$ and $x=7$, the $x$-coordinate of the center ($h$) is their average. The focus given is $(13, 5)$, so the $y$-coordinate of the center ($k$) must be 5.\r<BR>\n" );
document.write( "\n" );
document.write( "$$h = \frac{-5 + 7}{2} = \frac{2}{2} = 1$$<BR>\n" );
document.write( "$$\text{Center} (h, k) = \mathbf{(1, 5)}$$\r<BR>\n" );
document.write( "\n" );
document.write( "### B. Distance to Focus ($c$)\r<BR>\n" );
document.write( "\n" );
document.write( "The focus is at $(13, 5)$ and the center is at $(1, 5)$. The distance $c$ is the distance between them:<BR>\n" );
document.write( "$$c = 13 - 1 = \mathbf{12}$$\r<BR>\n" );
document.write( "\n" );
document.write( "### C. Distance between Directrices\r<BR>\n" );
document.write( "\n" );
document.write( "The distance between the directrices is $7 - (-5) = 12$. The general distance between the two directrices of a hyperbola is given by $2a/e$.\r<BR>\n" );
document.write( "\n" );
document.write( "$$2\frac{a}{e} = 12 \implies \frac{a}{e} = \mathbf{6} \quad \text{(Equation 1)}$$\r<BR>\n" );
document.write( "\n" );
document.write( "### D. Eccentricity ($e$) and Semi-transverse Axis ($a$)\r<BR>\n" );
document.write( "\n" );
document.write( "We use the relationship between $a, c,$ and the eccentricity $e$:<BR>\n" );
document.write( "$$c = a \cdot e$$<BR>\n" );
document.write( "Substituting $c=12$:<BR>\n" );
document.write( "$$12 = a \cdot e \quad \text{(Equation 2)}$$\r<BR>\n" );
document.write( "\n" );
document.write( "Now, we solve the system of equations (1) and (2) for $a$ and $e$:<BR>\n" );
document.write( "1. $a = 6e$ (from Eq. 1)<BR>\n" );
document.write( "2. $12 = (6e)e$ (substitute into Eq. 2)<BR>\n" );
document.write( "$$12 = 6e^2$$<BR>\n" );
document.write( "$$e^2 = 2 \implies e = \sqrt{2}$$\r<BR>\n" );
document.write( "\n" );
document.write( "Now find $a$:<BR>\n" );
document.write( "$$a = 6e = 6\sqrt{2}$$<BR>\n" );
document.write( "$$a^2 = (6\sqrt{2})^2 = 36 \times 2 = \mathbf{72}$$\r<BR>\n" );
document.write( "\n" );
document.write( "### E. Semi-conjugate Axis ($b^2$)\r<BR>\n" );
document.write( "\n" );
document.write( "Use the standard hyperbolic relation $c^2 = a^2 + b^2$.<BR>\n" );
document.write( "$$12^2 = 72 + b^2$$<BR>\n" );
document.write( "$$144 = 72 + b^2$$<BR>\n" );
document.write( "$$b^2 = 144 - 72 = \mathbf{72}$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 2. Standard Form of the Equation\r<BR>\n" );
document.write( "\n" );
document.write( "Since the hyperbola is horizontal, the standard form is:<BR>\n" );
document.write( "$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$<BR>\n" );
document.write( "Substitute $h=1, k=5, a^2=72,$ and $b^2=72$:\r<BR>\n" );
document.write( "\n" );
document.write( "$$\frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} = 1$$\r<BR>\n" );
document.write( "\n" );
document.write( "---\r<BR>\n" );
document.write( "\n" );
document.write( "## 3. General Form of the Equation\r<BR>\n" );
document.write( "\n" );
document.write( "To convert to the general form ($Ax^2 + Bx + Cy^2 + Dy + E = 0$), multiply the entire equation by the common denominator, 72:\r<BR>\n" );
document.write( "\n" );
document.write( "$$72 \left( \frac{(x-1)^2}{72} - \frac{(y-5)^2}{72} \right) = 1 \times 72$$<BR>\n" );
document.write( "$$(x-1)^2 - (y-5)^2 = 72$$\r<BR>\n" );
document.write( "\n" );
document.write( "Expand the squared terms:<BR>\n" );
document.write( "$$(x^2 - 2x + 1) - (y^2 - 10y + 25) = 72$$\r<BR>\n" );
document.write( "\n" );
document.write( "Remove parentheses and rearrange:<BR>\n" );
document.write( "$$x^2 - 2x + 1 - y^2 + 10y - 25 = 72$$<BR>\n" );
document.write( "$$x^2 - y^2 - 2x + 10y - 24 = 72$$\r<BR>\n" );
document.write( "\n" );
document.write( "Move the constant term to the left side:<BR>\n" );
document.write( "$$x^2 - y^2 - 2x + 10y - 24 - 72 = 0$$<BR>\n" );
document.write( "$$x^2 - y^2 - 2x + 10y - 96 = 0$$\r<BR>\n" );
document.write( "\n" );
document.write( "The equation of the hyperbola in general form is:<BR>\n" );
document.write( "$$\mathbf{x^2 - y^2 - 2x + 10y - 96 = 0}$$ </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );