Question

Mixed term $xy$ is to be removed from the general equation of second degree $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0,$ one should rotate the axes through an angle $\theta ,$ given by $\tan 2\theta$ equal to

• A. $\frac{a-b}{2h}$
• B. $\frac{2h}{a+b}$
• C. $\frac{a+b}{2h}$
• D. $\frac{2h}{a-b}$

Answer 1

The correct option is D $\frac{2h}{a-b}$

Rotating the axes thought an angle $\theta$ ,we have

$x=X\cos \theta -Y\sin \theta$ and $y=X\sin \theta +Y\cos \theta .$

$\because f(x,y)=a{x}^2+2hxy+2b{y}^2.$

After rotation, new equation is

$F(x,y)=(a{\cos }^2\theta +2h\cos \theta \sin \theta +b{\sin }^2\theta )X^2+2[(b-a)\cos \theta \sin \theta +h({\cos }^2\theta -{\sin }^2\theta )]XY+(a{\sin }^2\theta -2h\cos \theta \sin \theta +b{\cos }^2\theta )y^2$

Now, coefficient of $XY=0,$ then we get

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