A) \[{{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\cos \left( {{\sin }^{-1}}\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right) \right) \right]\]
B) \[\frac{150}{2}=\frac{150\sqrt{3}}{2}-10t\]
C) \[\Rightarrow \]
D) \[10t=\frac{150(\sqrt{3}-1)}{2}\]
Correct Answer: A
Solution :
Since, 3x + y = 0 is a tangent to the circle with centre at (2,-1). \[\therefore \]Radius = Length of the perpendicular from (2, -1) on 3x + y = 0 Radius\[=\frac{6-1}{\sqrt{9+1}}=\frac{5}{\sqrt{10}}=\sqrt{\frac{5}{2}}\] So, the equation of the circle is \[{{(x-2)}^{2}}+{{(y+1)}^{2}}=\frac{5}{2}\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}=4x+2y+\frac{5}{2}=0\] The combined equation of the tangents drav from the origin to this circle is\[S{{S}_{1}}={{T}^{2}}\] where, \[S={{x}^{2}}+{{y}^{2}}-4x+2y+\frac{5}{2},\] \[{{S}_{1}}={{0}^{2}}+{{0}^{2}}-4\times 0+2\times 0+\frac{5}{2}\] \[=\frac{5}{2}\] and \[T=x(0)+y(0)-2(x+0)+(y+0)+\frac{5}{2}\] \[=-2x+y+\frac{5}{2}\] \[\therefore \]\[\left( {{x}^{2}}+{{y}^{2}}-4x+2y+\frac{5}{2} \right)\left( \frac{5}{2} \right)={{\left( -2x+y+\frac{5}{2} \right)}^{2}}\] \[\Rightarrow \]\[3{{x}^{2}}-8xy-3{{y}^{2}}=0\] \[\Rightarrow \]\[3x+y=0\]and \[x-3y=0\]
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