translation and rotation of a parabola

Question

I am trying to translate a parabola to the origin, rotate by T radians and then translate back to the original position. I can calculate the new X and Y vectors using matrix operations and the regress to find the coefficients, but instead I'm trying to calculate the coeffecients of the new parabola.

This is what I've tried, but final values are wrong, so there's a mistake somewhere. Please would someone tell me if finding the new coeffs is possible (and my substitution is wrong) or if there is a better method?

Y=AX^2+BX+C vertex at (h,k)
Y'=A(X-h)^2+B(X-h)+C-k  translate vertex to (0,0)
X'=X-h  
Y''=Y'cosT+X'sinT   Y''=cosT*(A(X-h)^2+B(X-h)+C-k)+sinT*(X-h)
X''=Y'sinT-X'cosT   

Y'''=Y''+k  final translated, rotated, translated parabola (i think this is wrong)
Y'''=cosT(A(X-h)^2+B(X-h)+C-k)+sinT(X-h)+k  substitute Y'' to plot with original X values
X'''=X+h    

Y'''=A'X^2+B'X+C'   simplify eqn to find new coeffs

Answer 1

If you rotate the original parabola by any angle other than a multiple of $\pi$, the resulting curve cannot be expressed in the form $y=f(x)$. If you start with the equation $$ax^2+bx+c-y=0\tag{1}$$ instead, though, and apply the transformations to that, you’ll end up with a quadratic in $x$ and $y$ that you can then solve for $y$ using your favorite method (complete the square, quadratic formula, &c). It should be a pretty straightforward matter to plot the new parabola from there.

In particular, you’re looking to rotate the parabola through some angle $\theta$ about its vertex $(h,k)=\left(-{b\over2a},c-{b^2\over4a}\right)$. This transformation can be achieved by making the following substitutions in equation $(1)$: $$\begin{align} x&\to(x-h)\cos\theta-(y-k)\sin\theta+h \\ y&\to(x-h)\sin\theta+(y-k)\cos\theta+k. \end{align} \tag{2}$$ These are just the end result of applying the three transformations that you’ve listed. After rearranging and collecting terms, you end up with an equation of the form $$(\alpha x+\beta y)^2+\gamma x+\delta y+\epsilon=0$$ which you can then solve for $y$ via the quadratic formula: $$y = -{\alpha\over\beta}x\pm\frac12\sqrt{4(\alpha\delta-\beta\gamma)x-4\beta\epsilon+{\delta^2\over\beta}}\tag{3}$$

Depending on how you’re plotting these curves, however, a simpler approach might be to generate points on the original parabola and then transform them individually. That is, if you have some parametrization $(x(t),y(t))$ of the parabola, the right-hand sides of $(2)$ give you a parametrization of the rotated parabola.

Another approach is to use the polar form of the equation of a parabola with axis of symmetry parallel to the $y$-axis: $$r={2p\over1-\sin\theta}.\tag{4}$$ This parabola has its focus at the origin, with $p$ the focal distance. You can use this to parametrize the original parabola as $$\begin{align} x=h+r\cos t&=h+2p{\cos t\over1-\sin t}\\ y=k+p+r\sin t&=k+p+2p{\sin \over1-\sin t}. \end{align} \tag{5}$$ Note that since the polar equation places the focus at the origin, we have to add $p$ to the $y$-coordinate. The rotated parabola is a straightforward modification to $(5)$. Changing the orientation is just a matter of adding to the angle $t$, but we also have to adjust for the rotation of the focus. For a counterclockwise rotation of $\alpha$, this ends up as $$\begin{align} x&=h-p\sin\alpha+2p{\cos\left(t+\frac\alpha2\right)\over1-\sin\left(t-\frac\alpha2\right)}\\ y&=k+p\cos\alpha+2p{\sin\left(t+\frac\alpha2\right)\over1-\sin\left(t-\frac\alpha2\right)}.\end{align} \tag{6}$$ Finally, to relate this back to the original equation, we have $p={1\over4a}$, $h=-{b\over2a}$ and $k=c-{b^2\over4a}$.

Answer 2

first of all "Sorry for my english" .......I do a problem in my blog about a rotation of a parabola,wich is rotate about the focus and the equation of the guideline is $ax+by+c=0$ $$$$ firts cleared $y$. $$y=\frac{-a}{b}x-c$$ we calculated the slope of the perpendicular straight line. $$m_{2}=-\frac{1}{m_{1}}=\frac{-1}{-a/b}=\frac{b}{a}$$ we have $$y=\frac{b}{a}x$$ we want the interception points. $$\frac{b}{a}x=\frac{-a}{b}x-c$$ $$x=\frac{-abc}{a^{2}+b^{2}}$$ now found $y$: $$y=\frac{b}{a}\frac{-abc}{a^{2}+b^{2}}=\frac{-b^{2}c}{a^{2}+b^{2}}$$ write this result like a vector: $$\vec{S}=\left\langle\frac{-abc}{a^{2}+b^{2}}, \frac{-b^{2}c}{a^{2}+b^{2}}\right\rangle=\frac{bc}{a^{2}+b^{2}}\langle-a,-b\rangle$$ the guideline vector is: $$\vec{V}=\left\langle x,-\frac{a}{b}x\right\rangle$$ norm of $\vec{V}$ is: $$\|\vec{V}\|=x\sqrt{1+(a/b)^2}=\frac{x}{b}\sqrt{a^{2}+b^{2}}$$ then, the unitary vector $\hat{V}$ is: $$\hat{V}=\frac{1}{\frac{x}{b}\sqrt{a^{2}+b^{2}}}\left\langle x,-\frac{a}{b}x\right\rangle=\frac{1}{ \sqrt{a^{2}+b^{2}}}\left\langle b,-a\right\rangle$$ now we calculated the proyection in the guideline: $$\hat{V}\cdot\hat{X}=\frac{1}{\sqrt{a^{2}+b^{2}}}\left\langle b,-a\right\rangle\cdot\langle x,y\rangle=\frac{1}{\sqrt{a^{2}+b^{2}}}\left(bx-ay\right)$$ now we multiply the result above to the unitary vector guideline: $$\vec{V}_{proy}=\frac{1}{\sqrt{a^{2}+b^{2}}}\left(bx-ay\right)\frac{1}{\sqrt{a^{2}+b^{2}}}\left\langle b,-a\right\rangle =\frac{(bx-ay)}{a^{2}+b^{2}}\langle b,-a\rangle$$ calculate now $\vec{R}$ wich is the sum of $\vec{S}$ and $\vec{V}_{proy}$ $$\vec{R}=\frac{bc}{a^{2}+b^{2}}\langle-a,-b\rangle + \frac{(bx-ay)}{a^{2}+b^{2}}\langle b,-a\rangle$$ finally we calculate the distance between $\vec{R}$ and $\vec{X}$: $$\vec{D}=\frac{bc}{a^{2}+b^{2}}\langle-a,-b\rangle + \frac{(bx-ay)}{a^{2}+b^{2}}\langle b,-a\rangle-\langle x,y\rangle$$ $$\vec{D}=\frac{1}{a^{2}+b^{2}}[ \langle-abc,-b^{2}c\rangle+\langle b^{2}x-aby,a^{2}y-abx\rangle-\langle a^{2}x+b^{2}x,a^{2}y+b^{2}y\rangle]$$ $$\vec{D}=\frac{1}{a^{2}+b^{2}}\langle-abc-a^{2}x-aby,-b^{2}c-b^{2}y-abx\rangle$$ $$\vec{D}=\frac{1}{a^{2}+b^{2}}\langle a(-bc-ax-by),b(-bc-by-ax)\rangle$$ $$\vec{D}=\frac{-(bc+ax+by)}{a^{2}+b^{2}}\langle a,b\rangle$$ we now the eccentricity of parabola is $1$, then we have: $$\sqrt{\vec{X}\cdot\vec{X}}=\sqrt{\vec{D}\cdot\vec{D}}$$ $$\vec{X}\cdot\vec{X}=\vec{D}\cdot\vec{D}$$ this is the same to: $$x^{2}+y^{2}=\frac{(bc+ax+by)^{2}}{(a^{2}+b^{2})^{2}}(a^{2}+b^{2})$$ $$x^{2}+y^{2}=\frac{1}{a^{2}+b^{2}}(bc+ax+by)^{2}$$ $$x^{2}+y^{2}=\frac{1}{a^{2}+b^{2}}((bc)^{2}+2bc(ax+by)+(ax+by)^{2})$$ $$(a^{2}+b^{2})x^{2}+(a^{2}+b^{2})y^{2}=(bc)^{2}+2bcax+2b^{2}cy+a^{2}x^{2}+2abxy+b^{2}y^{2}$$ finally te equation of parabola is: $$b^{2}x^{2}-2abxy+a^{2}y^{2}-2bcax-2b^{2}cy=(bc)^{2}$$ if the parabola has a translation to the point $(h,k)$, then the equation is: $$b^{2}(x-h)^{2}-2ab(x-h)(y-k)+a^{2}(y-k)^{2}-2bca(x-h)-2b^{2}c(y-k)=(bc)^{2}$$ God bless you.

Answer 3

You can draw rotated parabola using pair of functions $$y_1=\frac{1}{4} \csc ^2(a) \left(2 \cos (a)-2 x \sin (2 a)- \sqrt{2+2 \cos (2 a)-16 x \sin (a)}\right)$$ and $$y_2=\frac{1}{4} \csc ^2(a) \left(2 \cos (a)-2 x \sin (2 a)+ \sqrt{2+2 \cos (2 a)-16 x \sin (a)}\right)$$ connected at point $$\left(\frac{1}{4} \cos (a) \cot (a),\frac{1}{4} (\cos (a)+\cot (a) \csc (a))\right)$$ where $a$ - angle of rotation. Sample Image.