document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=406437>Question 406437</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how to put : 5x^2+9y^2+1 into standard form for ellipse </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #286864 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=286864\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> how to put : 5x^2+9y^2+1 into standard form for ellipse\r<BR>\n" );
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document.write( "standard form for ellipse, x^2/a^2+y^2/b^2=1<BR>\n" );
document.write( "given equation :5x^2+9y^2+1<BR>\n" );
document.write( "=x^2/(1/5)+y^2/(1/9)=1\r<BR>\n" );
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document.write( "a^2=1/5<BR>\n" );
document.write( "a=sqrt(1/5)<BR>\n" );
document.write( "b^2=(1/9)<BR>\n" );
document.write( "b=(1/3)\r<BR>\n" );
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document.write( "Since the denominator,(1/5), of the x^2 term is greater than the denominator,(1/9), of the y^2 term, this is an ellipse with a horizontal major axis.Its center is at (0,0). </SPAN>\n" );
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