document.write( "Question 1142069: Find the coordinate of the vertices and foci and the equation of the asymptotes for the hyperbola with the equation
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Algebra.Com's Answer #762721 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
I'll only do one.  I'll do (v)\r\n" );
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document.write( "\"4x%5E2-25y%5E2-8x-96=0\"\r\n" );
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document.write( "Get terms in x² and x together.  Since there is no term in y, write y² as\r\n" );
document.write( "(y-0)²  \r\n" );
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document.write( "\"4x%5E2-8x-25%28y-0%29%5E2=96\"\r\n" );
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document.write( "Factor the coefficient of x² out of the x² and the x terms.\r\n" );
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document.write( "\"4%28x%5E2-2x%29-25%28y-0%29%5E2=96\"\r\n" );
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document.write( "Add and subtract 1 inside the first parentheses:\r\n" );
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document.write( "\"4%28x%5E2%5E%22%22-2x%2B1-1%29-25%28y-0%29%5E2=96\"\r\n" );
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document.write( "Factor the first three terms inside the first parentheses:\r\n" );
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document.write( "\"4%28%28x-1%29%5E2%5E%22%22-1%29-25%28y-0%29%5E2=96\"\r\n" );
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document.write( "Distribute the 4 into the big parentheses, removing the big parentheses,\r\n" );
document.write( "leaving the smaller parentheses intact:\r\n" );
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document.write( "\"4%28x-1%29%5E2-4-25%28y-0%29%5E2=96\"\r\n" );
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document.write( "Add 4 to both sides\r\n" );
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document.write( "\"4%28x-1%29%5E2-25%28y-0%29%5E2=100\"\r\n" );
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document.write( "Get 1 on the right side by dividing every term by 100\r\n" );
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document.write( "\"4%28x-1%29%5E2%2F100%5E%22%22-25%28y-0%29%5E2%2F100%5E%22%22=100%5E%22%22%2F100%5E%22%22\"\r\n" );
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document.write( "Simplify to get in standard form:\r\n" );
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document.write( "\"%28x-1%29%5E2%2F25%5E%22%22-%28y-0%29%5E2%2F4%5E%22%22=1\"\r\n" );
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document.write( "Since x comes first, the hyperbola opens left and right.\r\n" );
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document.write( "Compare to equation  \"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"\r\n" );
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document.write( "a² = 25, so a = 5;   b² = 4, so b = 2\r\n" );
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document.write( "Center = (h,k) = (1,0), semi-transverse axis = a = 5,\r\n" );
document.write( "semi-conjugate axis = b = 2    \r\n" );
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document.write( "Draw defining rectangle with center (h,k) = (1,0) with horizontal dimension\r\n" );
document.write( "equal to the entire transverse axis, or 2a = 2(5) = 10 units and vertical\r\n" );
document.write( "dimension equal to the the entire conjugate axis, or 2b = 2(2) = 4.\r\n" );
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document.write( "Draw and extend the diagonals of the defining rectangle.  They are the\r\n" );
document.write( "asymptotes.\r\n" );
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document.write( "Now we can sketch in the hyperbola.\r\n" );
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document.write( "The vertices are the midpoints of the left and right sides of the defining\r\n" );
document.write( "rectangle, (-4,0) and (6,0).\r\n" );
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document.write( "We calculate the foci, by using the Pythagorean relationship for all\r\n" );
document.write( "hyperbolas, which is c²=a²+b², where c is the distance from the center to\r\n" );
document.write( "the foci.\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
document.write( "\"c%5E2=5%5E2%2B2%5E2\"\r\n" );
document.write( "\"c%5E2=25%2B4\"\r\n" );
document.write( "\"c%5E2=29\"\r\n" );
document.write( "\"c=%22%22+%2B-+sqrt%2829%29\"\r\n" );
document.write( "\"c=%22%22+%2B-+5.4\", approximately\r\n" );
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document.write( "The word \"foci\" is the plural of the word \"focus\".\r\n" );
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document.write( "The left focus is c units left of the center (1,0) or \"%28matrix%281%2C3%2C1-sqrt%2829%29%2C%22%2C%22%2C0%29%29\"\r\n" );
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document.write( "The right focus is c units right of the center (1,0) or \"%28matrix%281%2C3%2C1%2Bsqrt%2829%29%2C%22%2C%22%2C0%29%29\"\r\n" );
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document.write( "They are the two black dots drawn below:\r\n" );
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document.write( "To find the equations of the asymptotes, they are the lines that go:\r\n" );
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document.write( "(1) through the center (1,0) and the upper right corner of the defining\r\n" );
document.write( "rectangle (6,2), which has slope 2/5, which is b/a, so the equation is\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-0=expr%282%2F5%29%28x-1%29\"\r\n" );
document.write( "\"y=expr%282%2F5%29%28x-1%29\"\r\n" );
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document.write( "(2) through the center (1,0) and the lower right corner of the defining\r\n" );
document.write( "rectangle (6,-2), which has slope -2/5, which is -b/a, so the equation is\r\n" );
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document.write( "\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"\r\n" );
document.write( "\"y-0=-expr%282%2F5%29%28x-1%29\"\r\n" );
document.write( "\"y=-expr%282%2F5%29%28x-1%29\"\r\n" );
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document.write( "Edwin
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