Removal of the term \[\mathbf{xy}\] from \[\mathbf{f(X,}\,\mathbf{Y)=a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+2hxy+b}{{\mathbf{y}}^{\mathbf{2}}}\] Without Changing the Origin
Category : JEE Main & Advanced
Clearly, \[h\ne 0\]. Rotating the axes through an angle \[\theta \], we have, \[x=X\cos \theta -Y\sin \theta \] and \[y=X\,\sin \theta +Y\cos \theta \]
\[\therefore \] \[f(x,y)=a{{x}^{2}}+2hxy+b{{y}^{2}}\]
After rotation, new equation is
\[F(X,Y)=(a{{\cos }^{2}}\theta +2h\cos \theta \sin \theta +b{{\sin }^{2}}\text{ }\theta ){{X}^{2}}\]
\[+2\{(b-a)\cos \theta \sin \theta +h({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )XY\]
\[+(a{{\sin }^{2}}\theta -2h\cos \theta \sin \theta +b{{\cos }^{2}}\theta ){{Y}^{2}}\]
Now coefficient of \[XY=0\]. Then we get \[\cot 2\theta =\frac{a-b}{2h}\].
Usually, we use the formula, \[\tan 2\theta =\frac{2h}{a-b}\] for finding the angle of rotation \[\theta \]. However, if \[a=b\], we use \[\cot 2\theta =\frac{a-b}{2h}\] as in this case \[\tan 2\theta \] is not defined.
You need to login to perform this action.
You will be redirected in
3 sec
