rotational of polynomials

Question

first of all it is well known that if we rotate (x,y) coordinate by some angle (let's say by A) then new image(x',y') will be related to (x,y) by the following formula x' = x*cosA - y*sin A and y' = x *sin A + y*cos A, now i have a question about this problem consider polynomial X^2-x*y+y^2=5,i know that there is rotation which eliminate xy term.(for second degree polynomials),source from which i am reading this problem says that it is a rotation by 45 degree and resulting equation after droping primes is 3*x^2+y^2=10,i am confused here how it is done?simple by the formula which i wrote above or there is another method?please help me

Answer 1

Indeed, the change of coordinates is calculated using a rotation matrix:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

You have the polynomial equation $P(x,y)=x^2-xy+y^2=5$. If we use a clockwise rotation of 45 degrees ($\theta=-45^{\circ}$ or $-\pi/4$ radians) on $(x,y)$ then the new coordinates will be

$$x' = \frac{x+y}{\sqrt{2}}, \quad y'=\frac{y-x}{\sqrt{2}},$$

because $\cos(-45^{\circ})=1/{\sqrt{2}}$ and $\sin(-45^{\circ})=-1/\sqrt{2}$. Plugging in the new values, we get

$$ P(x',y') = (x')^2 - x'y'+(y')^2=5 $$ $$\left(\frac{x+y}{\sqrt{2}}\right)^2 - \left(\frac{x+y}{\sqrt{2}}\right)\left(\frac{y-x}{\sqrt{2}}\right)+\left(\frac{y-x}{\sqrt{2}}\right)^2=5 $$

$$ \frac{x^2+2xy+y^2}{2}-\frac{y^2-x^2}{2}+\frac{y^2-2xy+x^2}{2}=5, $$

multiplying by 2, then after cancelling and grouping,

$$ 3x^2+y^2=10.$$

Hence the solution set $\{x',y'\}$ to the polynomial equation $P(x',y')=5$ can geometrically be visualized as the ellipse $\{(x,y):3x^2+y^2=10\}$ rotated clockwise by $45^{\circ}$:

$\hskip 2.2 in$