Rotation of ConicsDate: 12/01/1999 at 15:44:35 From: Joanna Schilling Subject: Rotation of Conics I have no idea where to start with this problem. Transform this into an equation with no xy term and graph the resulting equation. 17x^2 - 48xy + 31y^2 + 49 = 0 Date: 12/01/1999 at 16:22:41 From: Doctor Rob Subject: Re: Rotation of Conics Thanks for writing to Ask Dr. Math, Joanna. You can rotate coordinates by an angle t using the transformation x = cos(t)*x' - sin(t)*y' y = sin(t)*x' + cos(t)*y' The angle t that will make the x'y' term zero is given by tan(2*t) = b/(a-c), where a is the coefficient of x^2, b is the coefficient of x*y, and c is the coefficient of y^2. In this case, tan(2*t) = -48/(17-31) = 7/24 and then you can find sin(t) = 3/5 cos(t) = 4/5 The rest I leave to you. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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