Question

Let $\theta \in (0,\frac{\pi }{2})andx=Xcos\theta +ysin\theta ,y=Xsin\theta -Ycos\theta$ such that $x^2+2xy+y^2=a{X}^2+b{Y}^2$ Where a and b are constants then

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$x^2+y^2=x^2+y^2$
$xy=(x^2-y^2)sin\theta cos\theta -XY(co{s}^2\theta -si{n}^2\theta )$
$x^2+4xy+y^2=X^2+Y^2+2(X^2-Y^2)si{n}^2\theta -2XYco{s}^{2}theta$
=$1+2sin\theta )X^2+(1+2sin\theta )Y^2-2cos2\theta XY$
From the question
$a=1+2sin2\theta ,b=1-2sin2\theta ,cos2\theta =0$
$cos2\theta =0\Rightarrow \theta =\frac{\pi }{4},Thena=1+2sin\frac{\pi }{2}$
$b=1-2sin\frac{\pi }{2}$
$\therefore a=3,b=-1$