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
rotational of polynomials
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1Yes, 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. – 2011-07-24
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0i have tried it express x' and y' by the x,y and 45 degree but something is wrong – 2011-07-24
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1In 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 $A$... – 2011-07-24
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2"something is wrong" - then please show us your algebra and then we can help you from there. – 2011-07-24
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0so according formula x'=x* cos(45)-y*sin(45),y'=x*sin(45)+y*cos(45);now use formula x^2-x*y+y^2 and put instead of x and y ,use x' and y' yes?i have done so – 2011-07-24
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0Yes; now please write out what you have in your scratch paper so that we can be helpful... – 2011-07-24
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0first first if we take x' and y' in square and add we get x^2+y^2 if we multiply x' and y' we get (x^2)/2-(y^2)/2;now simple substract from x^2+y^2 our new result which is equal (x^2+3*y^2)/2 – 2011-07-24
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0@J.M. let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/909/discussion-between-user3196-and-j-m) – 2011-07-24
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0yes but i can't enter in chat room i am login in stackexchange but can't post messages in chat room – 2011-07-24
1 Answers
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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0i have made one very stupid mistake put x' and y' but forgot to make it equal 5 thank you very much @anon great explanation – 2011-07-24