A) \[{{\cos }^{-1}}\left[ \frac{4}{5} \right]\]
B) \[{{\sin }^{-1}}\left[ \frac{4}{5} \right]\]
C) \[{{\sin }^{-1}}\left[ \frac{3}{5} \right]\]
D) \[{{\tan }^{-1}}\left[ \frac{3}{5} \right]\]
Correct Answer: B
Solution :
Equation of pair of tangents to the circle \[{{x}^{2}}+{{y}^{2}}+4x+2y-4=0\]from the point (1, 1/2)is\[S{{S}_{1}}={{T}^{2}}\] \[({{x}^{2}}+{{y}^{2}}+4x+2y-4)\left( 1+\frac{1}{4}+4+1-4 \right)\] \[={{\left( x+\frac{1}{2}y+2(x+1)y+\frac{1}{2}-4 \right)}^{2}}\] \[\Rightarrow \]\[\frac{9}{4}({{x}^{2}}+{{y}^{2}}+4x+2y-4)={{\left( 3x+\frac{3}{2}y-\frac{3}{2} \right)}^{2}}\] \[\Rightarrow \]\[\frac{9}{4}{{x}^{2}}+\frac{9}{4}{{y}^{2}}+9x+\frac{9}{2}y-9\] \[=9{{x}^{2}}+\frac{9}{4}{{y}^{2}}+\frac{9}{4}+9xy-\frac{9}{2}y-9x\] \[\Rightarrow \] \[\frac{27}{4}{{x}^{2}}-18x-9y+9xy+\frac{45}{4}=0\] \[\Rightarrow \] \[\frac{3}{4}{{x}^{2}}+xy-2x-y+\frac{5}{4}=0\] \[\Rightarrow \] \[3{{x}^{2}}+4xt-8x-4y+5=0\] If\[\theta \]is the angle of these lines, then \[\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] \[\Rightarrow \] \[\tan \theta =\frac{2\sqrt{4}}{3}\] \[\Rightarrow \] \[\tan \theta =\frac{4}{3}\] \[\Rightarrow \] \[\sin \theta =\frac{4}{5}\] \[\Rightarrow \] \[\theta ={{\sin }^{-1}}\left[ \frac{4}{5} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
