1

How to rotate a conic by an determined angle? Could someone give me the step by step? (I know how to rotate the coordinate system by that formula \begin{align} x &= x'\cos(a) - y'\sin(a) \\ y &= x'\sin(a) + y'\cos(a) \end{align} but I don't know how to do it) Example: I want to rotate the hyperbola $x^2 - y^2 = 1$ by an angle of $45$, that is I want to find the equation of this hyperbola rotated by $45$ degrees in the original coordinate system ($xy$, not in $x'y'$)

David K
  • 98,388
danilocn94
  • 429

2 Answers2

0

So the formula in the rotated frame is:

\begin{equation}x'^2-y'^2=1\end{equation}

As you say, the rotation is defined as follows (also see here)

\begin{eqnarray} x' &=& x\cos(\theta)-y\sin(\theta) \\ y' &=& x\sin(\theta)+y\cos(\theta) \end{eqnarray}

So, simple substitution yields (noting that $\cos(45^{\circ})=\sin(45^\circ)=\sqrt{2}/2$):

\begin{eqnarray}(x-y)^2-(x+y)^2 &=& 2 \\ xy=-\frac{1}{2} \end{eqnarray}

Dr Xorile
  • 1,407
0

You can use matrices. You can express

$$Ax^2 + Bxy +Cy^2 +Fx +Gy +H=0,$$

as

$$\text{x}^TA\text{x} + J^T\text{x} + H=0,$$ where $$A = \begin{pmatrix}A & \frac{1}{2}B \\ \frac{1}{2}B & C \end{pmatrix}, \quad J = \begin{pmatrix}F\\ G\end{pmatrix}, \quad\text{x} = \begin{pmatrix} x \\ y\end{pmatrix}.$$

Then since $P = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}.$ is orthogonal and has $\det P = 1$, the transformation gives

$$(P\text{x})^TAP\text{x} + J^T(P\text{x}) + H=0,$$

Wortel
  • 205