Question

Show that the axes are to be rotated through an angle of $\frac{1}{2}{\tan }^{-1}\left(\frac{2h}{a-b}\right)$ so as to remove the $xy$ term from the equation $a{x}^2+2hxy+b{y}^2=0$, if $a\ne b$ and through the angle $\pi /4$, if $a=b$.

Answer 1

Let the x and y axis be rotated by an angle is as shown below:

then, $x=x\cos \theta -y\sin \theta$

$y=x\sin \theta +y\cos \theta$, where (x, y) is the coordinates with respect to new coordinate axes.

Given: $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

Replace $x\to x\cos \theta -y\sin \theta$

$y\to x\sin \theta +y\cos \theta$

$\Rightarrow a(x\cos \theta -y\sin \theta {)}^2+2h(x\cos \theta -y\sin \theta \left)\right(x\sin \theta +y\cos \theta )+b(x\sin \theta +y\cos \theta {)}^2+2g(x\cos \theta -y\sin \theta )+2f(x\sin \theta +y\cos \theta )+c=0$

$\Rightarrow a(x^2{\cos }^2\theta -y^2{\sin }^2\theta )+2h(x^2\sin \theta \cos \theta +xy{\cos }^2\theta -xy{\sin }^2\theta -y^2\sin \theta \cos \theta )+b(x^2{\sin }^2\theta +y^2{\cos }^2\theta )+2g(x\cos \theta -y\sin \theta )+2f(x\sin \theta +y\cos \theta )+c=0$

Now taking out every xy term

$-2axy\sin \theta \cos \theta +2hxy{\cos }^2\theta -2hxy{\sin }^2\theta +2hxy\sin \theta \cos \theta$

To eliminate the xy term, put coefficient of $xy=0$

$\Rightarrow -2a\sin \theta \cos \theta +(2h{\cos }^2\theta -2h{\sin }^2\theta )+2h\sin \theta \cos \theta =0$

$\Rightarrow 2h\left(\cos 2\theta \right)+\sin 2\theta (b-a)=0$

$\Rightarrow \tan 2\theta =\frac{2h}{a-b}$

$\Rightarrow \theta =\frac{1}{2}{\tan }^{-1}\left(\frac{2h}{a-b}\right)$.