Find equation of curve, by rotation of ellipse ?

**abrahamtim** · Mar 8th 2011, 01:55 AM

Find the equation of the curve obtained by rotation of the ellipse
$x^2 + 4y^2 = 4,$
by π/3 clock wise around the point (1, 2).

**earboth** · Mar 10th 2011, 11:15 PM

Originally Posted by abrahamtim

Find the equation of the curve obtained by rotation of the ellipse
$x^2 + 4y^2 = 4,$
by π/3 clock wise around the point (1, 2).

A) I hope this reply doesn't come too late ...

B) I'm not very practiced in mapping - so there must be a more elegant way to do your question.

1. Since the rotation has to be done clockwise the angle of rotation is $\alpha = -\frac{\pi}3$

2. I assume that you know how to rotate the coordinate system by:

$\left|\begin{array}{l}x = x' \cdot \cos(\alpha)-y' \cdot \sin(\alpha) \\ y = x' \cdot \sin(\alpha)+y' \cdot \cos(\alpha) \end{array}\right.$

where x', y' are the new coordinates after the rotation.

3. With $\sin\left(-\frac{\pi}3\right)=-\frac12 \cdot \sqrt{3}$ and $\cos\left(-\frac{\pi}3\right)=\frac12$
you'll get:

$x'=\frac12 (x-y\cdot \sqrt{3})$

$y'=\frac12 (x\cdot \sqrt{3} + y)$

Replace the variables x, y in the original equation by x', y' and you have rotated the ellipse around the origin.

4. The new midpoint of the ellipse is $M' \left(\frac12 - \sqrt{3},\ 1+\frac12 \cdot \sqrt{3}\right)$

5. Now translate the rotated ellipse by $\vec t = \left\langle\frac12 - \sqrt{3},\ 1+\frac12 \cdot \sqrt{3}\right\rangle$ and you'll get:

$13x^2 + 6 \sqrt{3} \cdot xy + x(20 \sqrt{3} - 22) + 7y^2 + 2y(2 - 5 \sqrt{3}) = 12 \sqrt{3} - 25$

6. I've attached a sketch.

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