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I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$

I tried to plot this but none of the graphing softwares that I use would allow it.

Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?

$$\left( \begin{array}{cc} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\ \end{array} \right)\left( \begin{array}{cc} x\\ y\\ \end{array} \right)=\left( \begin{array}{cc} X\\ Y\\ \end{array} \right)$$

For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$

$$\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \end{array} \right)\left( \begin{array}{cc} x\\ y\\ \end{array} \right)=\left( \begin{array}{cc} X\\ Y\\ \end{array} \right)$$

$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$ $$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$ $$y=x^2$$ $$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$ $$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$ $$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$ $$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$

Have I made a mistake somewhere?

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    $\begingroup$ Try desmos.com/calculator $\endgroup$ – Shuri2060 Jul 18 '17 at 19:25
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    $\begingroup$ It looks like you ended up rotating anticlockwise? $0=x^2+y^2-2xy-x\sqrt{2}-y\sqrt{2}$ is probably the right answer. desmos.com/calculator/4fkthwcvvd $\endgroup$ – Shuri2060 Jul 18 '17 at 19:28
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    $\begingroup$ desmos.com/calculator/mupzvz0i1i ... your equation is fine ... might have told us you started with $x=y^2$ ? $\endgroup$ – Donald Splutterwit Jul 18 '17 at 19:31
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    $\begingroup$ if $(1,-1)$ and $(1,1)$ were solutions prior to rotation, and you rotate 90 degrees clockwise, then $(\sqrt 2,0), (0,\sqrt 2)$ should be solutions post rotation. $\endgroup$ – Doug M Jul 18 '17 at 19:31
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    $\begingroup$ In general, if points are transformed according to some function $\phi$, you need to transform the equation by the inverse of $\phi$. E.g., to move the vertex of the parabola to $(h,k)$, you subtract these coordinates from the variables $x$ and $y$ in the equation. Similarly, to rotate the graph clockwise, you need to apply a counterclockwise rotation to the variables in the equation. $\endgroup$ – amd Jul 18 '17 at 21:04
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Let us start with general conic section

$$Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$$

or equivalently, we can write it as

$$\begin{pmatrix} x & y & 1 \end{pmatrix}\begin{pmatrix} A & B/2 & D/2\\ B/2 & C & E/2\\ D/2 & E/2 & F \end{pmatrix}\begin{pmatrix} x\\ y\\ 1 \end{pmatrix}=0$$

(we will denote the above 3x3 matrix with $M$)

So, let's say you are given a conic section $v^\tau M v = 0$ and let's say we want to rotate it by angle $\varphi$. We can represent appropriate rotation matrix with

$$Q_\varphi=\begin{pmatrix} \cos \varphi& -\sin\varphi & 0\\ \sin\varphi & \cos\varphi & 0\\ 0 & 0 & 1\end{pmatrix}$$

Now, $Q_\varphi$ represents anticlockwise rotation, so we might be tempted to write something like $$(Q_\varphi v)^\tau M (Q_\varphi v) = 0$$ to get conic section rotated by angle $\varphi$ anticlockwise. But, this will actually produce clockwise rotation. Think about it - if $v$ should be a point on the rotated conic, then $Q_\varphi v$ is a point on conic before rotation, thus, the last equation actually means that the new conic rotated anticlockwise will produce the old conic.

So, let us now do your exercise. You have conic $y = x^2$, so matrix $M$ is given by $$ M =\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & -1/2\\ 0 & -1/2 & 0\end{pmatrix}$$ and you want to rotate your conic clockwise by $\pi/4$, so choose $$Q_{\pi/4}=\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4 & 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}.$$

Finally, we get equation $$\begin{pmatrix} x & y & 1 \end{pmatrix}\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4 & 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}^\tau\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & -1/2\\ 0 & -1/2 & 0\end{pmatrix}\begin{pmatrix} \cos \frac\pi4& -\sin\frac\pi4& 0\\ \sin\frac\pi4 & \cos\frac\pi4 & 0\\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix} x\\ y\\ 1 \end{pmatrix}=0$$

or simplified $$x^2-2xy+y^2-x\sqrt 2-y\sqrt 2 = 0.$$

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  • $\begingroup$ Ennar can you have a look at my workings above? There must be a mistake somewhere that I can't find. $\endgroup$ – Kantura Jul 18 '17 at 20:54
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    $\begingroup$ @Derek, take a look at the paragraph where I say that $Q_\varphi$ seemingly produces anticlockwise rotation, but actually produces clockwise. I'm a bit tired and going for bed, so I'm not really reading into details that you wrote, but I think that is what you did. Does it make sense for you? $\endgroup$ – Ennar Jul 18 '17 at 20:56
  • $\begingroup$ Well I know that I rotated it anticlockwise but I don't know at which point I went wrong. $\endgroup$ – Kantura Jul 18 '17 at 21:02
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    $\begingroup$ @Derek, well I just told you, right. You represent points of the original conic with $(x,y)$, rotate it by $-\pi/4$ to get points of the new conic $(X,Y)$ and then you write $Y = X^2$ which is obviously wrong. Precisely the problem that I described in my answer. $\endgroup$ – Ennar Jul 18 '17 at 21:04
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    $\begingroup$ @Derek, what you should do is rotate $(X,Y)$ by $\pi/4$ to get $(x,y)$ (this is equivalent to rotating $(x,y)$ by $-\pi/4$ to get $(X,Y)$ like you did) and write $y = x^2$ instead. I've explained this in my answer, if you need clarification on it, let me know. $\endgroup$ – Ennar Jul 18 '17 at 21:08
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Your result is corect. This is the plot in geogebra.org

enter image description here

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    $\begingroup$ OP wanted clockwise rotation, not anticlockwise. $\endgroup$ – Ennar Jul 18 '17 at 20:51
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Before rotating (and plotting), you need to parametrize your parabola:

$$\begin{cases} x(t) = t\\ y(t) = t^2 \end{cases},$$

where $t \in \mathbb{R}.$ You are rotating each point of the parabola, and hence:

$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\cdot\begin{bmatrix}x(t)\\y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}x(t)-y(t)\\x(t)+y(t)\end{bmatrix}.$$

At the end you get that:

$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}t(1-t)\\t(1+t)\end{bmatrix}.$$

This can be plotted in Matlab using the following code:

t=linspace(-3,3,100);
X=(sqrt(2)/2)*(t.*(1-t));
Y=(sqrt(2)/2)*(t.*(1+t));
plot(X,Y)

This is what you get: enter image description here

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Rotating the parabola $y=x^2$ by $\theta$ clockwise gives $v=u^2$, where $$\left(u\atop v\right)=\left(\cos\theta\quad-\sin\theta\atop\sin\theta\quad\;\;\;\cos\theta\right)\left(x\atop y\right)$$ i.e. $$x\sin \theta+y\cos\theta=(x\cos\theta-y\sin\theta)^2$$ Putting $\theta=\frac\pi 4$ gives $$\frac 1{\sqrt2}(x+y)=\left(\frac 1{\sqrt2}(x-y)\right)^2\\ \sqrt2(x+y)=(x-y)^2$$ which when expanded is $$x^2+y^2-2xy-\sqrt2 x-\sqrt2 y=0$$

enter image description here

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  • $\begingroup$ But why do you use $\frac{\pi}{4}$ to go clockwise ? That is what confuses me. I mean if you go clockwise 45 degrees from the positive x-axis you will be at -45 degrees. So surely you should go $-\frac{\pi}{4}$ $\endgroup$ – Kantura Jul 19 '17 at 17:00
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    $\begingroup$ The matrix is just defined that way. The matrix you're looking for has the negative sign on the bottom left corner. You can derive this using the substitution x=cos(t) y = sin(t) making it an AC rotation $\endgroup$ – Dis-integrating Jul 19 '17 at 20:33
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    $\begingroup$ @Derek - Think of it as rotating the axes $\frac\pi 4$ anti-clockwise. $\endgroup$ – Hypergeometricx Jul 20 '17 at 2:05
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Maybe this is forcing unwarranted mathematics into the question but if you want to prove that the axis of symmetry rotates with the function then simply show that $M(\theta)R(\theta)(x,y)=R(\theta)(x,y)$ where $R$ and $M$ are rotation and reflection matrices respectively given that $\theta=0$ is the line of symmetry.

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