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Mathematics-Online course: Vector Calculus - Quadratic Curves

Rotation of Conic Sections

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If the conic section with the equation

$\displaystyle ax^2 + bxy + cy^2 + dx + ey +f = 0 $

is transformed by a rotation about the origin through the angle $ \alpha ,$ then the equation in the new coordinates $ \tilde{x}$ and $ \tilde{y}$ has the form

$\displaystyle \tilde{a}\tilde{x}^2 + \tilde{c}\tilde{y}^2 + \tilde{d}\tilde{x} + \tilde{e}\tilde{y} + \tilde{f} = 0\,, $

provided

$\displaystyle \tan 2\alpha = \frac{b}{a-c} \quad (a \neq c)$

or

$\displaystyle \alpha = \frac{\pi}{4} \quad (a = c)\ .$

The coordinate transformation which expresses the old coordinates in terms of the new ones is given by

$\displaystyle x= \tilde{x}\cos \alpha - \tilde{y}\sin \alpha \,,\quad y= \tilde{x}\sin \alpha + \tilde{y}\cos \alpha \,. $

The parameters of the new equation are

\begin{displaymath} \begin{array}{rclcrclcrcl} \tilde{a} &=& a \cos^2 \alpha +... ...ha - d \sin \alpha \,,&\ & \tilde{f} &=& f\,.\\ \end{array} \end{displaymath}

Note that in the new equation there is no mixed quadratic term, i.e. the coeffcient $ \tilde{b}$ of $ \tilde{x}\tilde{y}$ is zero. Thus this equation may be easily further transformed into a normal form of the conic section by completing squares.

Consider the conic section

$\displaystyle x^2 - 2 \sqrt{3}xy + 3y^2 - \sqrt{3}x - y + f = 0 .$

The condition on the angle $ \alpha $ of a suitable rotation of the coordinate axes is

$\displaystyle \tan 2\alpha = \sqrt{3} .$

Thus with $ \alpha = \frac{\pi}{6} $

the coordinate transformation

$\displaystyle x= \sqrt{3}\tilde{x}/2 - \tilde{y}/2 \,,\quad y= \tilde{x}/2 + \sqrt{3}\tilde{y}/2 $

yields

$\displaystyle \tilde{a}$ $\displaystyle =$ $\displaystyle 1\cdot 3/4 +(-2\sqrt{3}) \cdot \sqrt{3}/4 + 3\cdot 1/4 =0\,,$  
$\displaystyle \tilde{c}$ $\displaystyle =$ $\displaystyle 1\cdot 1/4 -(-2\sqrt{3}) \cdot \sqrt{3}/4+ 3\cdot 3/4 =4\,,$  
$\displaystyle \tilde{d}$ $\displaystyle =$ $\displaystyle -\sqrt{3} \cdot \sqrt{3}/2 +(-1)\cdot 1/2 = -2\,,$  
$\displaystyle \tilde{e}$ $\displaystyle =$ $\displaystyle -1 \cdot \sqrt{3}/2 -(-\sqrt{3})\cdot 1/2 = 0\,,$  

and the equation gets the form

$\displaystyle 4\tilde{y}^2 - 2\tilde{x} + f = 0 .$

Consequently the given conic section is a parabola (independent from $ f$ ).

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  automatically generated 10/30/2007