document.write( "Question 731952: Ellipse Using Standard Form\r
\n" ); document.write( "\n" ); document.write( "Vertices at (0,-1) and (12,-1), a focus at (6+√11-11). \n" ); document.write( "

Algebra.Com's Answer #447397 by josgarithmetic(39620) $\"\"$ $\"About$

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The focus given is not clear. Either the \"+\" is a comma or the \"-\" is a comma. No matter, if those vertices are the main vertices, then a start on the form of the equation is like, $\"%28x-6%29%5E2%2F36%2B%28y%2B1%29%5E2%2Fb%5E2=1\"$ , using center as half way between 0 and 12 on in reference to the x coordinate.\r
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\n" ); document.write( "\n" ); document.write( "CONTINUING TO ANALYZE:
\n" ); document.write( "Assuming your given vertices are on the major axis and are horizontally arranged, the y coordinate of the foci MUST be in the exact same horizontal line as the major axis. y=-1 MUST contain the foci. If you tried to say that the coordinate pair for one of the foci is (6+11^(1/2), 11), then this is wrong. Saying that one of the foci were (6+11^(1/2), -1) would make sense. If this were one of the foci, then since both foci are equally distant from the center, and x=6 is one of the coordinates of the center, then the other focus must be (6-11^(1/2), -1).
\n" ); document.write( "The way that a, b, and c are related is $\"a%5E2=b%5E2%2Bc%5E2\"$ . The value for c is $\"6%2B11%5E%281%2F2%29\"$ , which is the focal length. One simple algebra step gives $\"b%5E2=a%5E2-c%5E2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "The equation for the ellipse can be amended as
\n" ); document.write( " $\"%28x-6%29%5E2%2F36%2B%28y%2B1%29%5E2%2F%2836-%286%2B11%5E%280.5%29%29%5E2%29%5E2=1\"$ ,
\n" ); document.write( "and you can finish the arithmetic in that.\r
\n" ); document.write( "\n" ); document.write( "[The algebra.com system is not rendering this last equation very neatly. Read it carefully.] \n" ); document.write( "

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