A) \[2\sin {{18}^{o}}\]
B) \[2\sin {{15}^{o}}\]
C) \[\cos {{36}^{o}}\]
D) \[2\cos {{36}^{o}}\]
Correct Answer: A
Solution :
Tangent to the parabola \[{{y}^{2}}=4x\,\] is \[y=mx\,+\frac{1}{m}\] ?(i) \[\Rightarrow \,{{m}^{2}}x-my+1=0,\] As, it touches the circle \[{{x}^{2}}+{{y}^{2}}=1,\] so \[\left| \frac{1}{\sqrt{{{m}^{4}}+{{m}^{2}}}} \right|\,=1\Rightarrow \,{{m}^{4}}+{{m}^{2}}-1=0\] \[\therefore \,\,{{m}^{2}}={{\tan }^{2}}\theta \,=\frac{-1\pm \,\sqrt{1+4}}{2}=\frac{\sqrt{5}-1}{2}\] \[=2\left( \frac{\sqrt{5}-1}{4} \right)\,=2\sin {{18}^{0}}\]You need to login to perform this action.
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