document.write( "Question 1204148: Identify the co-vertices of the ellipse (x-3)^2 + {(y-5)^2/9} = 1 \n" ); document.write( "

Algebra.Com's Answer #840216 by Edwin McCravy(20060) $\"\"$ $\"About$

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document.write( "Learn to compare every ellipse equation with\r\n" );
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document.write( "If the larger denominator is under the x-term, then the ellipse looks like this  \r\n" );
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document.write( "If the larger denominator is under the y-term, then the ellipse looks like this  \r\n" );
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document.write( "Put a 1 under your first term:\r\n" );
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document.write( "So since the y-term has the larger denominator, you know it looks like this: \r\n" );
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document.write( "Write the denominators as squares:\r\n" );
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document.write( "The center is always the point (h,k).  'a' is always half of the major axis\r\n" );
document.write( "and 'b' is always half of the minor axis.  So your ellipse is of the form\r\n" );
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document.write( "'a' is always the square root of the larger denominator and 'b' is always the\r\n" );
document.write( "square of the smaller denominator. Again, your ellipse is of the form\r\n" );
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document.write( " compared to \r\n" );
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document.write( "The center is (h,k) = (3,5), a=3, b=1\r\n" );
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document.write( "So draw the center (3,5), the major axis goes up and down,\r\n" );
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document.write( "so draw in the major axis up 3 and down 3 from the center (3,5).\r\n" );
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document.write( "Then, draw in the minor axis left 1 and right 1 from the center (3,5). \r\n" );
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document.write( "Then draw in the ellipse:\r\n" );
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document.write( "Now you can look at the graph and see that the vertices are\r\n" );
document.write( "at the ends of the major axis, (3,2) and (3,8), and that the\r\n" );
document.write( "co-vertices are at the ends of the minor axis (2,5) and (4,5)\r\n" );
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document.write( "Edwin

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