Question

Let $0\le \alpha \le \frac{\pi }{2}$ and $x=X\cos \alpha +Y\sin \alpha ,y=X\sin \alpha -Y\cos \alpha$. lf $x^2+4xy+y^2=a{X}^2+b{Y}^2$, then

• A. $\alpha =\frac{\pi }{4}$
• B. $\alpha =\frac{\pi }{8}$
• C. $a+b=2$
• D. $a+b=4$

Answer 1

Answer: (A), (C)

The correct options are
A $\alpha =\frac{\pi }{4}$
C $a+b=2$

$x^2+y^2=X^2+Y^2$

$xy=(X^2-Y^2)\sin \alpha \cdot \cos \alpha -XY({\cos }^2\alpha -{\sin }^2\alpha )$
$\Rightarrow 4xy=(X^2-Y^2)2\sin 2\alpha -4XY\cos 2\alpha$
$\therefore x^2+y^2+4xy=X^2(1+2\sin 2\alpha )+Y^2(1-2\sin 2\alpha )-4XY\cos 2\alpha$
comparing with $x^2+y^2+4xy=a{X}^2+b{Y}^2$
we get
$\cos 2\alpha =0\Rightarrow \alpha =\frac{\pi }{4}$
$a=1+2\sin 2\alpha =3$
$b=1-2\sin 2\alpha =-1$
$\Rightarrow a+b=2$