A) \[{{\tan }^{-1}}(3/7)\]
B) \[{{\tan }^{-1}}(\sqrt{3}/7)\]
C) \[{{\tan }^{-1}}(7/3)\]
D) \[{{\tan }^{-1}}(3/\sqrt{7})\]
Correct Answer: D
Solution :
Given, \[x\text{ }cos\text{ }\alpha +y\text{ }sin\text{ }\alpha =4\] is a tangent to \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\] \[\Rightarrow \] \[y=-x\cot \alpha +4\operatorname{cosec}\alpha \] Here, \[m=-\cot \alpha ,\,\,c=4\,cosec\alpha \] and \[{{a}^{2}}=25,\,{{b}^{2}}=9\] Now, the condition of tangent to ellipse is \[{{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}\]s \[16\,\text{cose}{{\text{c}}^{2}}\alpha =25{{\cot }^{2}}\alpha +9\] \[\Rightarrow \] \[\frac{16}{{{\sin }^{2}}\alpha }=\frac{25{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha }+9\] \[\Rightarrow \] \[16=25{{\cos }^{2}}\alpha +9{{\sin }^{2}}\alpha \] \[\Rightarrow \] \[9{{\sin }^{2}}\alpha +25-25{{\sin }^{2}}\alpha =16\] \[\Rightarrow \] \[-16\,{{\sin }^{2}}\alpha =-9\] \[\Rightarrow \] \[{{\sin }^{2}}\alpha =\frac{9}{16},\,\sin \alpha =3/4\] \[\Rightarrow \] \[{{\cos }^{2}}\alpha =\frac{7}{16}\] \[\Rightarrow \] \[{{\tan }^{2}}\alpha =9/7\] \[\Rightarrow \] \[\alpha ={{\tan }^{-1}}(3/\sqrt{7})\]
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