A) \[\frac{1}{\sqrt{2}}\tan \left( \frac{x}{2}+\frac{\pi }{8} \right)+C\]
B) \[-\frac{1}{\sqrt{2}}\tan \left( \frac{x}{2}+\frac{\pi }{8} \right)+C\]
C) \[\frac{1}{\sqrt{2}}\cot \left( \frac{x}{2}+\frac{\pi }{8} \right)+C\]
D) \[-\frac{1}{\sqrt{2}}\cot \left( \frac{x}{2}+\frac{\pi }{8} \right)+C\]
Correct Answer: D
Solution :
Length of tangent from the point (1, 2) to the circle\[{{x}^{2}}+{{y}^{2}}+x+y-4=0\]is \[\sqrt{1+4+1+2-4}=2\] Similarly, length of tangent from the point (1,2) to the circle \[3{{x}^{2}}+3{{y}^{2}}-x-y-\lambda =0\] or\[{{x}^{2}}+{{y}^{2}}-\frac{x}{3}-\frac{y}{3}-\frac{\lambda }{3}=0\]is \[\sqrt{1+4-\frac{1}{3}-\frac{2}{3}-\frac{\lambda }{3}}=\sqrt{4-\frac{\lambda }{3}}\] But given that ratio of lengths of tangents to two circles is\[4:3\]. \[\therefore \] \[\frac{2}{\sqrt{4-\frac{\lambda }{3}}}=\frac{4}{3}\] \[\Rightarrow \] \[2\sqrt{4-\frac{\lambda }{3}}=3\] On squaring, \[4\left( 4-\frac{\lambda }{3} \right)=9\] \[\Rightarrow \] \[4-\frac{\lambda }{3}=\frac{9}{4}\] \[\Rightarrow \] \[\frac{\lambda }{3}=4-\frac{9}{7}=\frac{16-9}{4}=\frac{7}{4}\] \[\Rightarrow \] \[\lambda =\frac{21}{4}\]
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