document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209669>Question 1209669</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the asymptote of the hyperbola from part (a) with positive slope.\r<BR>\n" );
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document.write( "Enter your answer as an equation of the form .\r<BR>\n" );
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document.write( "-4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849783 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=849783\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to find the asymptote with a positive slope:\r<BR>\n" );
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document.write( "1. **Simplify and rearrange the equation:**\r<BR>\n" );
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document.write( "   -4x² + y² - 2y = -3y² + 8x + 9y + 15<BR>\n" );
document.write( "   -4x² - 8x + 4y² - 11y - 15 = 0<BR>\n" );
document.write( "   -4(x² + 2x) + 4(y² - (11/4)y) = 15<BR>\n" );
document.write( "   -4(x² + 2x + 1) + 4 + 4(y² - (11/4)y + (121/64)) - 121/16 = 15<BR>\n" );
document.write( "   -4(x + 1)² + 4(y - 11/8)² = 15 - 4 + 121/16<BR>\n" );
document.write( "   -4(x + 1)² + 4(y - 11/8)² = 11 + 121/16<BR>\n" );
document.write( "   -4(x + 1)² + 4(y - 11/8)² = (176 + 121)/16<BR>\n" );
document.write( "   -4(x + 1)² + 4(y - 11/8)² = 297/16\r<BR>\n" );
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document.write( "2. **Divide to get the standard form:**\r<BR>\n" );
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document.write( "   -(x + 1)²/(297/64) + (y - 11/8)²/(297/64) = 1<BR>\n" );
document.write( "   (y - 11/8)²/(297/64) - (x + 1)²/(297/64) = 1\r<BR>\n" );
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document.write( "3. **Identify the center and the values of a and b:**\r<BR>\n" );
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document.write( "   The center of the hyperbola is (-1, 11/8).<BR>\n" );
document.write( "   a² = 297/64, so a = √(297/64) = (3√33)/8<BR>\n" );
document.write( "   b² = 297/64, so b = √(297/64) = (3√33)/8\r<BR>\n" );
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document.write( "4. **Find the equations of the asymptotes:**\r<BR>\n" );
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document.write( "   The equations of the asymptotes for a hyperbola in the form (y-k)²/a² - (x-h)²/b² = 1 are given by:\r<BR>\n" );
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document.write( "   y - k = ±(a/b)(x - h)\r<BR>\n" );
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document.write( "   In our case:<BR>\n" );
document.write( "   y - 11/8 = ±((3√33)/8 / (3√33)/8)(x + 1)<BR>\n" );
document.write( "   y - 11/8 = ±1(x + 1)<BR>\n" );
document.write( "   y = ±(x+1) + 11/8\r<BR>\n" );
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document.write( "5. **Asymptote with positive slope:**\r<BR>\n" );
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document.write( "   We want the asymptote with the positive slope, so we take the positive sign:\r<BR>\n" );
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document.write( "   y = x + 1 + 11/8<BR>\n" );
document.write( "   y = x + 19/8\r<BR>\n" );
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document.write( "So, the equation of the asymptote with a positive slope is y = x + 19/8.<BR>\n" );
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