Algebra.Com's Answer #58918 by Edwin McCravy(20060)![\"\"](\"/images/stars/stars-13.gif\")  
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What is the standard form and state whether it is a parabola, a circle, an ellipse, or a hyperbla f 3x^2=8-4y^2-8y
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document.write( "First get a 0 on the right side:\r\n" );
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document.write( " 3x² + 4y² + 8y - 8 = 0 \r\n" );
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document.write( "Rule: If the polynomial equation of degree 2 contains no terms in xy \r\n" );
document.write( "and has 0 on the right side, then\r\n" );
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document.write( " I. If it has only one squared letter, it is a parabola.\r\n" );
document.write( " A. If the squared letter is x², the axis of symmetry is vertical\r\n" );
document.write( " B. If the squared letter is y², the axis of symmetry is horizontal.\r\n" );
document.write( "II. If it has two squared letters then\r\n" );
document.write( " A. If the coefficients of the squared letters have the same sign\r\n" );
document.write( " then it is an ellipse.\r\n" );
document.write( " 1. If the coefficients of the squared letters are also equal, it is a \r\n" );
document.write( " circle, which is a special kind of ellipse.\r\n" );
document.write( " 2. If the coefficients of the squared letters are not equal but have the\r\n" );
document.write( " same sign, it is an ellipse which is not a circle.\r\n" );
document.write( " 3. If the coefficient of x² is smaller in absolute value than the\r\n" );
document.write( " coefficient of y², then its major axis is horizontal. \r\n" );
document.write( " 4. If the coefficient of y² is smaller in absolute value than the\r\n" );
document.write( " coefficient of x², then its major axis is vertical. \r\n" );
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document.write( " B. If the coefficients of the squared letters have opposite signs, then\r\n" );
document.write( " it is a hyperbola.\r\n" );
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document.write( "So\r\n" );
document.write( " 3x² + 4y² + 8y - 8 = 0\r\n" );
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document.write( "fits II A 2, so it is an ellipse which is not a circle, and whose\r\n" );
document.write( "major axis is horizontal since 3 < 4.\r\n" );
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document.write( "To get it in standard form, we have to make it look like this:\r\n" );
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document.write( "  +  =  \r\n" );
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document.write( "Get the constant term on the right:\r\n" );
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document.write( " 3x² + 4y² + 8y = 8 \r\n" );
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document.write( "Since there are no terms in x, we complete the square on x simply\r\n" );
document.write( "by replacing x by (x - 0) \r\n" );
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document.write( " 3(x - 0)² + 4y² + 8y = 8\r\n" );
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document.write( "To complete the square on the y terma, first factor out the\r\n" );
document.write( "coefficient of y² out of both the y² and the y terms:\r\n" );
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document.write( " 3(x - 0)² + 4(y² + 2y) = 8\r\n" );
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document.write( "Multiply the coefficient of y, which is 2, by 1/2, getting 1\r\n" );
document.write( "Then square the 1 getting 1. Add this inside the parentheses\r\n" );
document.write( "on the left. This amounts to adding 4×1 to the left side of\r\n" );
document.write( "the equation since the parentheses has a 4 multiplied by it.\r\n" );
document.write( "Therefore to compensate, we must add 4 to the right side:\r\n" );
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document.write( " 3(x - 0)² + 4(y² + 2y + 1) = 8 + 4\r\n" );
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document.write( "Factor the trinomial inside the second parentheses as a\r\n" );
document.write( "perfect square, combine terms on the right:\r\n" );
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document.write( " 3(x - 0)² + 4(y + 1)² = 12\r\n" );
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document.write( "Get 1 on the right side by dividing every term by 12\r\n" );
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document.write( "  +  =  \r\n" );
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document.write( "Divide top and bottom of the first fraction by 3, top and bottom\r\n" );
document.write( "of the second fraction by 4 and simplify the right side as just 1:\r\n" );
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document.write( "  +  = \r\n" );
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document.write( "That's the standard form., which is all you're looking for. But\r\n" );
document.write( "you will eventually be asked for more, so here is some more:\r\n" );
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document.write( "Comparing that to:\r\n" );
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document.write( " + = \r\n" );
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document.write( "h = 0, k = -1, a = 2, b =  \r\n" );
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document.write( "So its center is (h,k) = (0,-1). Its semi-major akis is a = 2, and\r\n" );
document.write( "its semi-minor axis is b = \r\n" );
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document.write( "To draw its graph we locate the center, and draw the complete major\r\n" );
document.write( "axis and minor axes intersecting at the center (0,-1), like this:\r\n" );
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document.write( "Then sketch in the ellipse:\r\n" );
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document.write( "It looks almost like a circle, but if you'll look carefully it's\r\n" );
document.write( "not really because its a little bit fatter than it is tall. They \r\n" );
document.write( "only begin to look egg-shaped when b is less than half of a.\r\n" );
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document.write( "The two foci (focal points) are on the major axis each a distance\r\n" );
document.write( " of c units from the center, where c² = a² - b²\r\n" );
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document.write( " c² = 4 - 3 \r\n" );
document.write( " c² = 1 \r\n" );
document.write( " c = 1\r\n" );
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document.write( "So the foci are at (1,1) and (-1,-1) marked below as o's,\r\n" );
document.write( "1 unit from the center (0,-1)\r\n" );
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document.write( "Edwin \r
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