question_answer

A line L is common tangent to the circle \[{{x}^{2}}+{{y}^{2}}=1\] and the parabola\[{{y}^{2}}=4x\]. If \[\theta \] is the angle which it makes with the positive x-axis, then \[{{\tan }^{2}}\theta \] is equal to

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 \[\left| \frac{1}{\sqrt{{{m}^{4}}+{{m}^{2}}}} \right|\,=1\Rightarrow \,{{m}^{4}}+{{m}^{2}}-1=0\] \[\therefore \,\,{{m}^{2}}={{\tan }^{2}}\theta \,=\frac{-1\pm \,\sqrt{1+4}}{2}=\frac{\sqrt{5}-1}{2}\] \[=2\left( \frac{\sqrt{5}-1}{4} \right)\,=2\sin {{18}^{0}}\]