Question

By shifting the origin to the point $(2,3)$ and then rotating the coordinate axes through an angle $\theta$ in the counter clockwise direction, if the equation
$3 x^{2}+2 x y+3 y^{2}-18 x-22 y+50=0$ is transformed to $4 X^{2}+2 Y^{2}-1=0$, then the angle $\theta=$

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (C)

By shifting the origin to the point $(2,3)$ and then rotating the coordinate
axes through an angle $\theta$ in counter clockwise direction, the
$x=2+X \cos \theta-Y \sin \theta$
and $y=3+x \sin \theta+y \cos \theta$, so the given equation
$3 x^{2}+2 x y+3 y^{2}-18 x-22 y+50=0$
is transformed to
$3(2+x \cos \theta-y \sin \theta)^{2}+2(2+x \cos \theta-y \sin \theta)$
$(3+x \sin \theta+y \cos \theta)$
$+3(3+x \sin \theta+y \cos \theta)^{2}-18(2+x \cos \theta$
$-y \sin \theta)-22(3+x \sin \theta+y \cos \theta)+50=0$
and given as $4 x^{2}+2 y^{2}-1=0$ and as $x y$ is missing so
$\tan 2 \theta=\frac{2}{3-3} $
$\Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$