If for $ heta \in\left[-\frac{\pi}{3}, 0 ight]$, the points $(x, y)=\left(3 an \left( heta+\frac{\pi}{3} ight), 2 an \left( heta+\frac{\pi}{6} ight) ight)$ lie on $x y+\alpha x+\beta y+\gamma=0$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Question

If for $	heta \in\left[-\frac{\pi}{3}, 0ight]$, the points $(x, y)=\left(3 	an \left(	heta+\frac{\pi}{3}ight), 2 	an \left(	heta+\frac{\pi}{6}ight)ight)$ lie on $x y+\alpha x+\beta y+\gamma=0$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} & x=3\left(\frac{	an 	heta+\sqrt{3}}{1-\sqrt{3} 	an 	heta}ight) \ & x-\sqrt{3} 	an 	heta=3 	an 	heta+3 \sqrt{3} \ & 	an 	heta=\frac{x-3 \sqrt{3}}{3+\sqrt{3} x} \ldots(1) \ & 2\left(\frac{	an 	heta+\frac{1}{\sqrt{3}}}{1-\frac{	an 	heta}{\sqrt{3}}}=yight) \ & 2(\sqrt{3} 	an 	heta+1)=y(\sqrt{3}-	an 	heta) \ldots . \end{aligned}$ using (1) and (2) $\begin{aligned} & 2\left(\frac{x-3 \sqrt{3}}{\sqrt{3}+x}+1ight)=y\left(\sqrt{3}-\frac{(x-3 \sqrt{3})}{\sqrt{3}(\sqrt{3}+x)}ight) \ & 2 \sqrt{3}(x-3 \sqrt{3}+x+\sqrt{3})=y(3(\sqrt{3}+x)-x+3 \sqrt{3}) \ & 4 \sqrt{3} x-12=y(2 x+6 \sqrt{3}) \ & x y-2 \sqrt{3} x+3 \sqrt{3} y-6=0 \ & \Rightarrow \alpha=-2 \sqrt{3}, \beta=3 \sqrt{3}, \gamma=-6 \ & \alpha^2+\beta^2+\gamma^2=12+27+36=75 \end{aligned}$ ^

If for $\theta \in\left[-\frac{\pi}{3}, 0\right]$, the points $(x, y)=\left(3 \tan…

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