Reducing $x^2+xy+2y^2=4$ to the equation of a conic

Question

I'm trying to reduce $x^2+xy+2y^2=4$ to the equation of a conic by rotating the axes through a clockwise roation of $a$ using the following change of variables: $$x=X\cos(a) + Y\sin(a)$$ $$y=-X\sin(a) + Y\cos(a)$$ where X and Y are the rotated axes. After making the change of variables I arrived at: $$X^2(1+\sin^2(a)-\frac{1}{2}\sin(2a))+Y^2(1+\cos^2(a)+\frac{1}{2}\sin(2a))+XY(-\sin(2a)+\cos(2a))=4$$ Now, to remove the $XY$ term I chose $a=\frac{\pi}{8}$, however since there isn't an exact value associated with $\frac{\pi}{8}$ for $\sin$ or $\cos$ then I can't find an exact value for the coefficients of $X^2$ and $Y^2$, which is what I'm supposed to have. So I'n wondering if there is angle I can choose or something else I can do to remove the $XY$ term and also have an exact value for the coefficients of $X$ and $Y$.

Answer 1

In the development of $$(cx+sy)^2+(cx+sy)(-sx+cy)+2(-sx+cy)^2,$$

the $xy$ terms have the coefficient

$$2cs+(c^2-s^2)-4cs=c_2-2s_2.$$

It is cancelled when

$$t_2=\frac12=\frac{2t}{1-t^2}.$$

Then

$$t^2+4t-1=0$$ yields $t=\sqrt5-2$ and $t^2=9-4\sqrt5$.

The remaining polynomial is

$$(c^2-cs+2s^2)x^2+(s^2+sc+2c^2)y^2.$$

Factoring out $c^2$,

$$c^2((1-t+2t^2)x^2+(t^2+t+2)y^2)=4,$$

$$c^2((21-9\sqrt5)x^2+(9-3\sqrt5)y^2)=4.$$

Finally, $$c^2=\frac1{1+t^2}=\frac{2\sqrt5+5}{10}.$$

Answer 2

COMMENT.-You have chosen badly the coordinates; in fact do you have $$x=X\cos(a)-Ysin(a)\\y=X\sin(a)+Y\cos(a)$$ where $X,Y$ are the new axes.

The equation $x^2+xy+2y^2-4=0$ seen as a general equation of the second degree $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is such that $A\ne C$ and both have same sign then one has an ellipse whose center is the origin (Why?) (option: the discriminant $B^2-4AC\lt 0$ so one has an ellipse).You want a reduced form of this ellipse changing axes by rotation.

Let $\mathscr C$ be the conic. Note that $(\dfrac 12,\dfrac54)\in\mathscr C$ and its opposite $(-\dfrac 12,-\dfrac54)\in\mathscr C$ and the corresponding perpendicular passing by the origin gives the points $(\dfrac{10}{\sqrt{23}},\dfrac{-4}{\sqrt{23}})$ and $(\dfrac{-10}{\sqrt{23}},\dfrac{4}{\sqrt{23}})$ in $\mathscr C$.

Calculations given finally $$\frac{23X^2}{116}+\frac{16Y^2}{29}=1$$

Can you now calculate the rotation angle?