Q83 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
When the coordinate axes are rotated through an angle $\theta$ in anti clockwise direction, if the transformed equation of $x^2+y^2+2 x y+2 x+6 y+1=0$ is
$(2+\sqrt{3}) X^2+2 X Y+(2-\sqrt{3}) Y^2+a X+b Y+2=0$
then $3 a-b=$
- A10
- B$2(1+2 \sqrt{3})$
- Correct answerC20
- D$2(3+\sqrt{3})$
Solution
If coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction, then
$x=X \cos \theta-Y \sin \theta$
and $y=X \sin \theta+Y \cos \theta$, on substituting the values of $x$ and $y$ in equation
$x^2+y^2+2 x y+2 x+6 y+1=0 \text {, we get }$
$(X \cos \theta-Y \sin \theta)^2+(X \sin \theta+Y \cos \theta)^2$ $+2(X \cos \theta-Y \sin \theta)(X \sin \theta+Y \cos \theta)$ $+2(X \cos \theta-Y \sin \theta)+6(X \sin \theta+Y \cos \theta)+1=0$
$\Rightarrow\left(\cos ^2 \theta+\sin ^2 \theta+2 \sin \theta \cos \theta\right) X^2$ $+\left(-2 \sin \theta \cos \theta+2 \sin \theta \cos \theta+2 \cos ^2 \theta-2 \sin ^2 \theta\right) X Y$ $+\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta\right) Y^2$ $+(2 \cos \theta+6 \sin \theta) X+(6 \cos \theta-2 \sin \theta) Y+1=0$
$\Rightarrow(1+\sin 2 \theta)+2 \cos 2 \theta X Y+(1-\sin 2 \theta) Y^2$ $+(2 \cos \theta+6 \sin \theta) X+(6 \cos \theta-2 \sin \theta) Y+1=0$
On comparing with transformed given equation
$(2+\sqrt{3}) X^2+2 X Y+(2-\sqrt{3}) Y^2+a X+b Y+2=0,$
we get
$\sin 2 \theta=\frac{\sqrt{3}}{2}, \cos 2 \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{6}$
$\therefore \quad a=4(\cos \theta+3 \sin \theta)=2 \sqrt{3}+6$
and $b=4(3 \cos \theta-\sin \theta)=6 \sqrt{3}-2$
$\therefore 3 a-b=6 \sqrt{3}+18-6 \sqrt{3}+2=20$
Hence, the correct answer is option: 'C'.