A) \[{{\tan }^{-1}}\sqrt{2}\]
B) \[{{\tan }^{-1}}2\]
C) \[{{\tan }^{-1}}2\sqrt{2}\]
D) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: C
Solution :
Given equation of curve is \[{{y}^{2}}=4x\] ??.(i) and \[{{x}^{2}}+{{y}^{2}}=12\] ?...(ii) Let the slope of curve (i) is m^ and curve (ii) is \[({{m}_{2}})\]. Then, from Eqs. (i) and (ii), \[2y\frac{dy}{dx}=4\Rightarrow {{m}_{1}}=\frac{2}{y}\] and \[2x+2y\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{m}_{2}}=\frac{-x}{y}\] Let the angle between curve at intersection point is \['\theta '\]. \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}.{{m}_{2}}} \right|=\left| \frac{2/y-(-x/y)}{1+(2/y)\,(-x/y)} \right|\] \[\tan \theta =\left| \frac{\frac{2}{y}+\frac{x}{y}}{1-\frac{2x}{{{y}^{2}}}} \right|=\left| \frac{(2+x)y}{{{y}^{2}}-2x} \right|\] ??(iii) On solving Eqs. (i) and (ii), we get \[{{x}^{2}}+4x=12\] \[\Rightarrow \] \[{{x}^{2}}+4x-12=0\] \[\Rightarrow \] \[{{x}^{2}}+6x-2x-12=0\] \[\Rightarrow \] \[(x+6)\,(x-2)=0\] \[\Rightarrow \] \[x=-6,2\] (here \[x\ne -6\]) \[\Rightarrow \] \[y=\pm \,\,2\sqrt{2}\] So, the intersection point \[=(2,\,\pm \,\,2\sqrt{2})\] From Eq. (iii) \[\tan \theta =\left| \frac{(2+2)\,\,(\pm \,2\sqrt{2})}{8-4} \right|=\left| 4\frac{(\pm \,2\sqrt{2})}{4} \right|\] \[\tan \theta =|\pm \,\,2\sqrt{2}|=2\sqrt{2}\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\,\,(2\sqrt{2})\]
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