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EAMCET
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Mathematics
The transformed equation of $x^2+6xy+8y^2=10$ when the axes are rotated through an angle $\large\frac{\pi}{4}$ is :
\[\begin {array} {1 1} (a)\;15x^2-14xy+3y^2=20 & \quad (b)\;15x^2+14xy-3y^2=20 \\ (c)\;15^2+14xy+3y^2=20 & \quad (d)\;15x^2-14xy-3y^2=20 \end {array}\]
jeemain
eamcet
math
2006
q44
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asked
Nov 6, 2013
by
meena.p
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1 Answer
$ (c)\;15^2+14xy+3y^2=20$
answered
Nov 7, 2013
by
pady_1
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