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Rotation of Conics

Date: 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   
Associated Topics:
College Conic Sections/Circles
College Trigonometry

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