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
$3x^2+2xy+3y^2-18x-22y+50=0$ is transformed to $4X^2+2Y^2-1=0$, 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
$3x^2+2xy+3y^2-18x-22y+50=0$
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 $4x^2+2y^2-1=0$ and as $xy$ is missing so
$\tan 2\theta =\frac{2}{3-3}$
$\Rightarrow 2\theta =\frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}$