Question

The equation of a pair of straight lines is ${ax}^2+2hxy+b{y}^2=0$. By what angle must the axes be rotated so that the term containing $xy$ in the equation may be removed?

• A. $\theta ={\tan }^{-1}\frac{a}{b}$
• B. $\theta ={\tan }^{-1}\left(\frac{2h}{a-b}\right)$
• C. $\theta =\frac{1}{2}{\tan }^{-1}\left(\frac{2h}{a-b}\right)$
• D. None of these

Answer 1

Answer: (C)

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

Substitute $x=X\cos \theta -Y\sin \theta$ and $y=X\sin \theta +Y\cos \theta$ in the equation,

$a{(X\cos \theta -Y\sin \theta )}^2+2h(X\cos \theta -Y\sin \theta )(X\sin \theta +Y\cos \theta )+b{(X\sin \theta +Y\cos \theta )}^2=0$
Coefficient of $xy=-2a\cos \theta \sin \theta +2b\cos \theta \sin \theta +2h({\cos }^2\theta -{\sin }^2\theta )=0$
$\Rightarrow (b-a)\sin 2\theta +2h\cos 2\theta =0$
$\Rightarrow \tan 2\theta =\frac{2h}{a-b}$

$\Rightarrow \theta =\frac{1}{2}{\tan }^{-1}\frac{2h}{a-b}$