A) 1
B) \[\sqrt{3}\]
C) 3
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: C
Solution :
Solving equations \[{{x}^{2}}+{{y}^{2}}=5\] and \[{{y}^{2}}=4x\] we get \[{{x}^{2}}+4x-5=0\] i.e., \[x=1,\,-5\] For \[x=1\]; \[{{y}^{2}}=4\] Þ \[y=\pm 2\] For \[x=-5\]; \[{{y}^{2}}=-20\](imaginary values) \ Points are (1, 2)(1, ?2); \[{{m}_{1}}\] for \[{{x}^{2}}+{{y}^{2}}=5\] at (1, 2) \[{{\left. \frac{dy}{dx}=-\frac{x}{y} \right|}_{(1,\,2)}}=-\frac{1}{2}\]Similarly, \[{{m}_{2}}\] for \[{{y}^{2}}=4x\] at (1,2) is 1. \\[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{-\frac{1}{2}-1}{1-\frac{1}{2}} \right|=3\].
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