# rotational of polynomials

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

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Yes, you simply use the formula that you gave. Write the first equation with primes. Substitute the rotation formulas. Then, choose the angle of the coefficient of the resulting $x\cdot y$ term so that the coefficient is zero. – Raskolnikov Jul 24 '11 at 9:33
i have tried it express x' and y' by the x,y and 45 degree but something is wrong – dato datuashvili Jul 24 '11 at 9:38
In particular, the $xy$ "cross term" in your equation should become something like $xy\cos\,2A+\frac12(x^2-y^2)\sin\,2A$ post-substitution. You then pick the appropriate $A$... – J. M. Jul 24 '11 at 9:41
"something is wrong" - then please show us your algebra and then we can help you from there. – J. M. Jul 24 '11 at 9:42
so according formula x'=x* cos(45)-ysin(45),y'=xsin(45)+ycos(45);now use formula x^2-xy+y^2 and put instead of x and y ,use x' and y' yes?i have done so – dato datuashvili Jul 24 '11 at 9:45

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 $45^{\circ}$:

$\hskip 2.2 in$

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i have made one very stupid mistake put x' and y' but forgot to make it equal 5 thank you very much @anon great explanation – dato datuashvili Jul 24 '11 at 10:08