document.write( "Question 1082607: What is the equation of the following graph?
\n" ); document.write( " ellipse through the points (1, negative 2), (0, 1), (negative 1, negative 2) and (0, negative 5)
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Algebra.Com's Answer #696658 by natolino_2017(77)\"\" \"About 
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let (x-h)^2/a^2 + (y-k)^2/b^2 = 1. the general equation.\r
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\n" ); document.write( "\n" ); document.write( "1) (1-h)^2/a^2 + (2+k)^2/b^2 = 1
\n" ); document.write( "2) h^2/a^2 + (1-k)^2/b^2 =1
\n" ); document.write( "3) (1+h)^2/(a^2) + (2+k)^2/b^2 = 1
\n" ); document.write( "4) h^2/a^2 + (5+k)^2/b^2 = 1\r
\n" ); document.write( "\n" ); document.write( "3) - 1) 4h/a^2 =0 so h = 0.\r
\n" ); document.write( "\n" ); document.write( "4) - 2) (24 + 12k)/b^2 = 0, so k =-2.\r
\n" ); document.write( "\n" ); document.write( "replacing on 1) 1/a^2 = 1, so a^2 = 1.\r
\n" ); document.write( "\n" ); document.write( "replacing on 2) 9/b^2 = 1, so b^2 = 9.\r
\n" ); document.write( "\n" ); document.write( "Answer: x^2 + (y+2)^2/9 = 1 \r
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