Question

If by rotating the axes through an angle $\theta$ the general equation of second degree $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ is free from the term of xy, then prove that $\tan 2\theta$ is $\frac{2h}{a-b}$.

Answer 1

Let the axes be rotated through an angle $\theta$
so that $\begin{array}{c}x\to x\cos \theta -y\sin \theta \\ y\to x\sin \theta +y\cos \theta \end{array}\}$
$a(x\cos \theta -y\sin \theta {)}^2+2h(x\cos \theta -y\sin \theta \left)\right(x\sin \theta +y\cos \theta )$

$+b(x\sin \theta +y\cos \theta {)}^2+2g(x\cos \theta -y\sin \theta )+2f(x\sin \theta +y\cos \theta )+c=0$
Collect the coefficients of xy and put it equal to zero
$(-a\sin 2\theta +b\sin 2\theta )+2h({\cos }^2\theta -{\sin }^2\theta )=0$
$\frac{\sin 2\theta }{\cos 2\theta }=\frac{2h}{a-b}\Rightarrow \tan 2\theta =\frac{2h}{a-b}$.