Q83 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics

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
    $\frac{\pi}{2}$
  • Correct answer
    C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{3}$

Solution

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 $$ \begin{aligned} & 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 \end{aligned} and given as $4 x^2+2 y^2-1=0$ and as $x y$ is missing so $\tan 2 \theta=\frac{2}{3-3} \Rightarrow 2 \theta=\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}$

Hence, the correct answer is option: 'C'.

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