document.write( "Question 1209667: Parts (a)-(c) refer to the graph of
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document.write( "-4x^2 + y^2 - 2y = -3y^2 + 8x + 9y + 15
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document.write( "which is a hyperbola.\r
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document.write( "Find the center of the hyperbola. \n" );
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Algebra.Com's Answer #849785 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Here's how to find the center of the hyperbola:\r \n" ); document.write( "\n" ); document.write( "1. **Simplify and rearrange the equation:**\r \n" ); document.write( "\n" ); document.write( " -4x² + y² - 2y = -3y² + 8x + 9y + 15 \n" ); document.write( " -4x² - 8x + 4y² - 11y - 15 = 0 \n" ); document.write( " -4(x² + 2x) + 4(y² - (11/4)y) = 15 \n" ); document.write( " -4(x² + 2x + 1) + 4 + 4(y² - (11/4)y + (121/64)) - 121/16 = 15 \n" ); document.write( " -4(x + 1)² + 4(y - 11/8)² = 15 - 4 + 121/16 \n" ); document.write( " -4(x + 1)² + 4(y - 11/8)² = 11 + 121/16 \n" ); document.write( " -4(x + 1)² + 4(y - 11/8)² = (176 + 121)/16 \n" ); document.write( " -4(x + 1)² + 4(y - 11/8)² = 297/16\r \n" ); document.write( "\n" ); document.write( "2. **Divide to get the standard form:**\r \n" ); document.write( "\n" ); document.write( " -(x + 1)²/(297/64) + (y - 11/8)²/(297/64) = 1 \n" ); document.write( " (y - 11/8)²/(297/64) - (x + 1)²/(297/64) = 1\r \n" ); document.write( "\n" ); document.write( "3. **Identify the center:**\r \n" ); document.write( "\n" ); document.write( " The standard form of a hyperbola centered at (h, k) is:\r \n" ); document.write( "\n" ); document.write( " (y - k)²/a² - (x - h)²/b² = 1 (for a vertical hyperbola)\r \n" ); document.write( "\n" ); document.write( " Comparing this with our equation, we can see that:\r \n" ); document.write( "\n" ); document.write( " h = -1 \n" ); document.write( " k = 11/8\r \n" ); document.write( "\n" ); document.write( "Therefore, the center of the hyperbola is (-1, 11/8). \n" ); document.write( " \n" ); document.write( " |