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The origin is shifted to the point $(2,3)$ by translation of axes and then the coordinate axes are rotated…
The origin is shifted to the point $(2,3)$ by translation of axes and then the coordinate axes are rotated about the origin through an angle $\theta$ in the counter-clockwise sense. Due to this 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 $\theta=$
$\frac{\pi}{4}$ $\frac{\pi}{3}$ $\frac{\pi}{6}$ $\frac{\pi}{2}$ Solution
$3 x^2+2 x y+3 y^2-18 x-22 y+50=0$
Shifting origin to the point $(2,3)$, put $x=X+2, y=Y+3$
$\begin{aligned}
& 3(X+2)^2+2(X+2)(Y+3)+3(Y+3)^2 \\
& \quad-18(X+2)-22(Y+3)+50=0 \\
& \Rightarrow 3 X^2+2 X Y+3 Y^2-1=0
\end{aligned}$
Now, for rotating about $\theta$ angle, put
$\begin{aligned}
& X=x^{\prime} \cos \theta-y^{\prime} \sin \theta \text { and } Y=x^{\prime} \sin \theta+y^{\prime} \cos \theta \\
& \therefore 3\left(x^{\prime} \cos \theta-y^{\prime} \sin \theta\right)^2+2\left(x^{\prime} \cos \theta-y^{\prime} \sin \theta\right) \\
& \left(x^{\prime} \sin \theta+y^{\prime} \cos \theta\right)+3\left(x^{\prime} \sin \theta+y^{\prime} \cos \theta\right)^2-1=0 \\
& \Rightarrow\left(3 \cos ^2 \theta+3 \sin ^2 \theta+2 \sin \theta \cos \theta\right) x^{\prime 2} \\
& +\left(3 \sin ^2 \theta+3 \cos ^2 \theta-2 \sin \theta \cos \theta\right) y^{12} \\
& +\left(2 \cos ^2 \theta-2 \sin ^2 \theta-6 \sin \theta \cos \theta\right. \\
& +6 \sin \theta \cos \theta) x^{\prime} y^{\prime}-1=0 \\
& \Rightarrow(3+\sin 2 \theta) x^{\prime 2}+(3-\sin 2 \theta) y^{\prime 2}+(2 \cos 2 \theta) x^{\prime} y^{\prime}-1=0
\end{aligned}$
Comparing with $4 x^2+2 y^2-1=0$ we get,
$2 \cos 2 \theta=0 \Rightarrow \theta=\frac{\pi}{4}$
Asked in: AP EAMCET 2024 (23 May Shift 1)
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