document.write( "Question 716150: Given the equation (x+3)^2/36+(y-1)^2/18=1
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document.write( "find: the center. I found (-3,1)
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document.write( "vertices (-3.-5) and (-3,7)
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document.write( "the end points of minor axis I only came up with 6 radical 2
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document.write( "the foci I got 3 radical 2 and length of major axis is 12
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document.write( "Are these correct?
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document.write( "I can't find the length of focal cord or its endpoints.
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document.write( "Please help \n" );
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Algebra.Com's Answer #439757 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You got the center in the right place. But the vertices are at the ends of the major axis. With the absence of an \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "As for the endpoints of the minor axis, take the square root of the smaller denominator. Since 18 is 9 times 2, the square root of 18 is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The semi-major axis is 6, so the major axis has to be 12.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The measure of the semi-latus rectum (i.e. half of the focal chord) is given by \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The latus rectum, or focal chord if you will, is perpendicular to the major axis at the focus. So for your ellipse, the left hand focal chord has end points (-9,4) and (-9,-2). I'll leave the endpoints of the other focal chord in your capable hands.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "Egw to Beta kai to Sigma \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |