# Math Help - Rotation of a conic?

1. ## Rotation of a conic?

I am having some difficulty answer this type of question.

x^2-5x+y^2=3 rotated by pi/3

cos pi/3= 1/2 sin pi/3= sqrt{3}/2

But I am unsure how to proceed my book says

x with x cos theta + y sin theta

y with -x sin theta + y cos theta

Then plug this in the original equation

I think the graph is a circle

2. The transformed point of $(x,y)$ by means of the rotation is

$\begin{Bmatrix} x'=\dfrac{1}{2}x-\dfrac{\sqrt{3}}{2}y\\{}\\y'=\dfrac{\sqrt{3}}{2}x+ \dfrac{1}{2}y\end{matrix}$

Equivalently

$\begin{Bmatrix} x=\dfrac{1}{2}x'+\dfrac{\sqrt{3}}{2}y'\\y=-\dfrac{\sqrt{3}}{2}x'+\dfrac{1}{2}y'\end{matrix}$

Now, substitute in the original conic.