Rotate the parabola $y=x^2$ clockwise $45^\circ$.

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

Your result is corect. This is the plot in geogebra.org

Answer 4

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:

Answer 5

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 $$x^2+y^2-2xy-\sqrt2 x-\sqrt2 y=0$$