document.write( "Question 65288: If the eccentricity of an ellipse (in other words, the value e) is close to zero, then the ellipse looks almost like a circle.
\n" ); document.write( "True
\n" ); document.write( " False \r
\n" ); document.write( "\n" ); document.write( "Each of the remaining questions refer to the ellipse 4x^ 2 + 9y ^2 - 8x + 36y + 4 = 0.
\n" ); document.write( "The center of the ellipse is at the point
\n" ); document.write( " (-1,2)
\n" ); document.write( " (-2,1)
\n" ); document.write( " (1,-2)
\n" ); document.write( " (2,-1)
\n" ); document.write( "The foci of the ellipse are the points \r
\n" ); document.write( "\n" ); document.write( "The length of the major axis of the ellipse is
\n" ); document.write( "6 units
\n" ); document.write( " 3 units
\n" ); document.write( " 4 units
\n" ); document.write( " 2 units
\n" ); document.write( "The length of the minor axis is
\n" ); document.write( "3 units
\n" ); document.write( " 6 units
\n" ); document.write( " 4 units
\n" ); document.write( " 2 units
\n" ); document.write( " \r
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Algebra.Com's Answer #45854 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
1). The eccentricity of an ellipe is given by:
\n" ); document.write( "\"e+=+c%2Fa\" Where c is the distance of the foci from the centre of the ellipse and a is half the length of the major axis. If c is close to zero, the foci are close to the centre and, therefore, the ellipse is close to being a circle.\r
\n" ); document.write( "\n" ); document.write( "2). Given the equation of the ellipse:
\n" ); document.write( "\"4x%5E2%2B9y%5E2-8x%2B36y%2B4+=+0\", find the coordinates of the centre, the length of the major axis, and the length of the minor axis.
\n" ); document.write( "To find these you should get your equation into the standard form for an ellipse with its centre at (h, k), semi-major axis = a and semi-minor axis = b.
\n" ); document.write( "\"%28%28x-h%29%5E2%29%2Fa%5E2+%2B+%28%28y-k%29%5E2%29%2Fb%5E2+=+1\" and to do this you will need to use the process of \"completing the square\". But first rearrange your equation as follows:
\n" ); document.write( "\"%284x%5E2-8x%29+%2B+%289y%5E2%2B36y%29+%2B+4+=+0\" Subtract 4 from both sides.
\n" ); document.write( "\"%284x%5E2-8x%29+%2B+%289y%5E2+%2B+36%29+=+-4\" Now complete the squares in the x-terms and in the y-terms by making the coefficients of the \"x%5E2\" and the \"y%5E2\" term equal to 1. You can do this by dividing through by LCD of 36.
\n" ); document.write( "\"%284x%5E2-8x%29%2F36+%2B+%289y%5E2%2B36y%29%2F36+=+-4%2F36\" Now cancel where appropriate.
\n" ); document.write( "\"%28x%5E2-2x%29%2F9+%2B+%28y%5E2%2B4y%29%2F4+=+-1%2F9\" Now complete the squares in the x-terms and the y-terms by adding \"1%2F9\" and \"4%2F4\" respectively to both sides of the equation.
\n" ); document.write( "\"%28x%5E2-2x%2B1%29%2F9+%2B+%28y%5E2%2B4y%2B4%29%2F4+=+-1%2F9+%2B+1%2F9+%2B+4%2F4\" Simplifying this, you get.
\n" ); document.write( "\"%28%28x-1%29%5E2%29%2F3%5E2+%2B+%28%28y%2B2%29%5E2%29%2F2%5E2+=+1\" Compare this with the standard form:
\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1\" with centre at (h, k), a = 3 and b = 2\r
\n" ); document.write( "\n" ); document.write( "So the centre of the ellipse is at: (1, -2)
\n" ); document.write( "The foci are located on the major axis at c units from the centre, where:
\n" ); document.write( "\"c%5E2+=+a%5E2+%2B+b%5E2\"
\n" ); document.write( "\"c+=+sqrt%289%2B4%29\"
\n" ); document.write( "\"c+=+sqrt%2813%29\"
\n" ); document.write( "The major axis is parallel to the x-axis, so the foci are located at:
\n" ); document.write( "(1+sqrt(13), -2) and (1-sqrt(13), -2)\r
\n" ); document.write( "\n" ); document.write( "The length of the major axis is 2a = 2(3) = 6.
\n" ); document.write( "The length of the minor axis is 2b = 2(2) = 4.
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