Question

Let $0\le \theta \le \frac{\pi }{2}$ and $x=X\cos \theta +Y\sin \theta ,y=X\sin \theta -Y\cos \theta$ such that $x^2+2xy+y^2=a{X}^2+b{Y}^2$, where $a$ and $b$ are constant then:

• A. $a=2,b=0$
• B. $\theta =\frac{\pi }{2}$
• C. $a=3,b=-1$
• D. $\theta =\frac{\pi }{3}$

Answer 1

The correct option is A $a=2,b=0$

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

$x^2=X^2{\cos }^2\theta +Y^2{\sin }^2\theta +2XY\sin \theta \cos \theta$

$y^2=X^2{\sin }^2\theta +Y^2{\cos }^2\theta -2XY\sin \theta \cos \theta$

$2xy=2(X\cos \theta +Y\sin \theta )(X\sin \theta -Y\cos \theta )$

$=2(X^2\sin \theta \cos \theta -XY{\cos }^2\theta +XY{\sin }^2\theta -Y^2\sin \theta \cos \theta )$

$=X^2\sin 2\theta -2XY\cos 2\theta -Y^2\sin 2\theta$

or $x^2+y^2+2xy$

$=X^2{\sin }^2\theta +Y^2{\cos }^2\theta -2XY\sin \theta \cos \theta +X^2{\cos }^2\theta +Y^2{\sin }^2\theta +2X^4\sin \theta \cos \theta +X^2\sin 2\theta -2XY\cos 2\theta -Y^2\sin 2\theta$

$=X^2+Y^2+X^2\sin 2\theta -2XY\cos 2\theta -Y^2\sin 2\theta$

$=X^2(1+\sin 2\theta )+Y^2(1-\sin 2\theta )-2XY{\cos }^2\theta$

acc of question

$X^2(1+\sin 2\theta )+Y^2(1-\sin 2\theta )-2XY\cos 2\theta =a{X}^2+b{Y}^2$

Compare

$2XY\cos 2\theta =0$

$\cos 2\theta =0$

$\cos 2\theta =\cos \pi /2$

$\theta =\frac{\pi }{4}$

Put the values of $\theta$

$X^2(1+\sin 2\frac{\pi }{4})+Y^2(1-\sin 2.\frac{\pi }{4})=0=a{X}^2+b{Y}^2$

$X^2(1+1)+Y^2(1-1)=a{X}^2+b{Y}^2$

$2X^2+o{Y}^2=a{X}^2+b{Y}^2$

$a=2$ $b=0$