# Axes Rotation Problem

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.

• See here. You will find many details on how to eliminate the xy term for conic sections. Commented Jan 21, 2016 at 2:37
• en.wikipedia.org/wiki/Rotation_of_axes Commented Jan 21, 2016 at 3:02

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.

• Can you please explain how you got 6sc and 4(c^2-s^2) Commented Jan 21, 2016 at 3:36
• Coefficient of $XY$ from the 1st term is $-2cs$, from the 2nd term is $4(c^2-s^2)$, from the penultimate term is $8sc$. Add them up and equate to $0$ and you get $6sc=4(c^2-s^2)$. Commented Jan 21, 2016 at 3:40
• Okay, thank you. So to figure out the new equation, do I just substitute s and c back into the long equation then simplify? Commented Jan 21, 2016 at 3:42
• Why did you divide the final equation by 5? Because I got 5Y^2+5X+10Y+1=0 Commented Jan 21, 2016 at 3:59
• Because I'm lazy and it's easier to write... Commented Jan 21, 2016 at 4:10

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ 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 $\theta=26.56505^\circ$