Axes Rotation Problem

Question

Given $$x^2 - 4xy + 5(\sqrt5y) + 4y^2 + 1 = 0$$ rotate the axes to eliminate the $xy$-term in the equation, then write the equation is standard form.

Answer 1

Suppose your new coordinates are $$ \begin{bmatrix} X\\ Y \end{bmatrix}=\begin{bmatrix} \cos\phi&\sin\phi\\ -\sin\phi&\cos\phi \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} $$ Let $c=\cos\phi,s=\sin\phi$. Your new equation is $$ (Xc-Ys)^2-4(Xc-Ys)(Xs+Yc)+5\sqrt5(Xs+Yc)+4(Xs+Yc)^2+1=0\\ $$ As there's no $XY$ term we have $$ 6sc=4(c^2-s^2)\implies \tan2\phi={4\over3}\implies \phi={1\over2}\arctan({4\over3})\\ \implies c={2\over\sqrt5},s={1\over\sqrt5} $$ The new equation becomes $$ Y^2+X+2Y+0.2=0 $$ So you have to rotate your axes by $\phi$ radian counter-clockwise.

Answer 2

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ Then $$\tan(2\theta)=\dfrac{B}{A–C}$$

For $x^2−4xy+5(\sqrt{5}y)+4y^2+1=0$, $\tan(2\theta)=\dfrac{4}{3}$

imply $2\theta=53.1301^\circ$ and the angle of rotation is $\theta=26.56505^\circ$