If for [-( )/(3),0], the points (x,y)=(3 ( +( )/(3)),2 ( +( )/(6))) lie on xy+ x+ y+ =0, t…

Question

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

Accepted Answer

$$\begin{aligned}
&\text { Let } \phi=\theta+\frac{\pi}{3} \Rightarrow \theta=\phi-\frac{\pi}{6}\
&\begin{aligned}
& x=3 \tan \left(\theta+\frac{\pi}{3}\right)=3 \tan \left(\theta+\frac{\pi}{6}\right) \
& y=2 \tan \phi \
& \tan \left(\phi+\frac{\pi}{6}\right)=\frac{\tan \phi+\frac{1}{\sqrt{3}}}{1-\tan \phi \cdot \frac{1}{\sqrt{3}}} \
& \frac{x}{3}=\frac{\frac{y}{2}+\frac{1}{\sqrt{3}}}{1-\frac{y}{2} \cdot \frac{1}{\sqrt{3}}}
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow \quad x=\frac{3(y \sqrt{3}+2)}{2 \sqrt{3}-y} \
& x y+\alpha x+\beta y+r=0 \
& 3\left(\frac{y \sqrt{3}+2}{2 \sqrt{3}-y}\right)+\alpha\left(3 \frac{(y \sqrt{3}+2)}{(2 \sqrt{3}-y)}\right)+\beta y+r=0 \
& =(3 \sqrt{3}-\beta) y^2+(6+3 \sqrt{3} \alpha+2 \sqrt{3} \beta-y) y \
& \quad+(6 \alpha+2 \sqrt{3} y)=0
\end{aligned}$$

For this identity to hold for all $\theta$, coefficients must be 0

$$\begin{aligned}
& \therefore \quad \beta=3 \sqrt{3} \
& \gamma=-\alpha \sqrt{3} \
& 6+3 \sqrt{3} \alpha+(2 \sqrt{3})(3 \sqrt{3})+\alpha \sqrt{3}=0 \
& \Rightarrow \alpha=-2 \sqrt{3} \
& \Rightarrow \beta=6 \
& \alpha^2+\beta^2+\gamma^2=75
\end{aligned}$$

Solutions · 1