document.write( "Question 385429: this is from the shifted conics sections, you have to identify whether it is a parabola, ellipse or hyperbola, and solve and graph the problem: 2x^2 + y^2 = 2y + 1 \n" ); document.write( "

Algebra.Com's Answer #272719 by Edwin McCravy(20059) $\"\"$ $\"About$

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document.write( "2x² + y² = 2y + 1\r\n" );
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document.write( "We place it in the form\r\n" );
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document.write( "Ax² + Bxy + Cy² + Dx + Ey + F = 0\r\n" );
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document.write( "2x² + 0xy + 1y² + 0x - 2y - 1 = 0\r\n" );
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document.write( "A = 2\r\n" );
document.write( "B = 0\r\n" );
document.write( "C = 1\r\n" );
document.write( "D = 0\r\n" );
document.write( "E = -2\r\n" );
document.write( "F = -1\r\n" );
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document.write( "The rules are:\r\n" );
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document.write( "1. if the discriminant, B² − 4AC < 0, the equation represents an ellipse; \r\n" );
document.write( "2. if A = C and B = 0, the equation represents a circle; \r\n" );
document.write( "3. if B2 − 4AC = 0, the equation represents a parabola; \r\n" );
document.write( "4. if B2 − 4AC > 0, the equation represents a hyperbola; \r\n" );
document.write( "5. if we also have A + C = 0, the equation represents a rectangular hyperbola. \r\n" );
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document.write( "The discriminant is B²-4AC = 0²-4(2)(1) = -8, so it represents an ellipse.\r\n" );
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document.write( "Since it has no xy term, we get it into one of these forms\r\n" );
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document.write( " or \r\n" );
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document.write( "Either way we begin by getting the constant on the right side:\r\n" );
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document.write( "2x² + y² - 2y = 1\r\n" );
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document.write( "We complete the square on y by adding 1 to both sides:\r\n" );
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document.write( "2x² + y² - 2y + 1 = 1 + 1\r\n" );
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document.write( "We factor the last three terms on the left as a perfect square:\r\n" );
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document.write( "2x² + (y - 1)² = 2\r\n" );
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document.write( "We divide through by 2 to get 1 on the right side:\r\n" );
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document.write( "We write x as (x-0)\r\n" );
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document.write( "Write the first term over 1\r\n" );
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document.write( "That is the form \r\n" );
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document.write( "because a is always greater than b in an ellipse.\r\n" );
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document.write( "It has center (h,k) = (0,1), semi-major axis = a = 1,\r\n" );
document.write( "and semi-minor axis = b = \r\n" );
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document.write( "It has its major axis vertical because a² is under the term in y\r\n" );
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document.write( "We plot the center:\r\n" );
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document.write( "We draw the complete major axis , about 1.4 units up and down\r\n" );
document.write( "from the center (in purple):\r\n" );
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document.write( "We draw the complete minor axis  unit right and left\r\n" );
document.write( "from the center (in green):\r\n" );
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document.write( "We sketch in the ellipse:\r\n" );
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document.write( "Edwin

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