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I have a parabola $y^2=4x$, and I have to rotate this by $120^\circ$ in the anticlockwise direction. What my book does is, it takes a point $(x,y)$ in the complex form, ie, $z=x+\iota y$, and then by rotation, the new point is $z_1=ze^{\frac{2\pi}{3}} = \bigg(-\frac{x}{2}-\frac{\sqrt3 y}{2}\bigg)+\iota \bigg(\frac{\sqrt3x}{2}-\frac{y}{2}\bigg)$

So the equation of the new curve, the book says is

$\bigg(\frac{\sqrt3x}{2}-\frac{y}{2}\bigg)^2=4\bigg(-\frac{x}{2}-\frac{\sqrt3 y}{2}\bigg)$

But the axis of this curve is $\sqrt3 x = y$, which has a slope of $\sqrt3$, ie it makes an angle of $60^\circ$ with the positive direction of the $x$-axis.

My question is, is rotating the point on the curve by an angle (as the book does), same as rotating the axis of a curve by that angle? If not, why? In what "sense" is this rotation?

Ted Shifrin
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Techie5879
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  • Rotating a parabola will obviously rotate all points, axis, directrix, generatrix, focus and whatnot. –  Apr 16 '20 at 15:10
  • @YvesDaoust By that same angle right? – Techie5879 Apr 16 '20 at 15:10
  • Same goes with any similarity transformation. –  Apr 16 '20 at 15:12
  • @Techie5879 Yes, all by the same angle, since rigidity is assumed. – Allawonder Apr 16 '20 at 15:16
  • @Allawonder Yes, so what they are doing seems perfectly sound to me but the graphing out of the equations doesn't make sense, it looks like the curve is rotated in a clockwise sense to me, although what they do algebraically seems perfectly sound to me. – Techie5879 Apr 16 '20 at 15:17

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I believe they may have mixed up the original and new co-ordinates in the book.

Just as, if you want to translate the graph of $y=f(x)$ by $a\in\mathbb R$ horizontally, the resulting graph has the equation $y=f(x-a)$ and not $y=f(x+a)$.

The correct solution would be not to express $z_1=x_1+iy_1=ze^{2\pi i/3}$, but to express $z=x+iy=z_1e^{-2\pi i/3}$, which gives you $x, y$ expressed via $x_1, y_1$, and then, the original equation $y^2=4x$, by substituting, gives you the equation of the rotated curve in terms of $x_1, y_1$.

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    Indeed: I was going to write the same answer. – Intelligenti pauca Apr 16 '20 at 15:18
  • But wouldn't rotating a curve once clockwise and anticlockwise for the same angle just undo the transformation? – Techie5879 Apr 16 '20 at 15:19
  • @Techie5879 It would indeed. The original curve is obtained by the opposite (clockwise) rotation from the new curve. We use that as we take the co-ordinates $(x_1, y_1)$ on the new curve, rotate them back, and what we get must satisfy the given equation for the original curve - which gives you the equation $x_1, y_1$ had to satisfy in the first place. –  Apr 16 '20 at 15:22
  • Yes thanks, this makes sense, they just put the points after 1 rotation in the old curve, that shouldn't really satisfy the old curve. Got it – Techie5879 Apr 16 '20 at 15:56