A) \[\frac{\pi }{2}\]
B) \[{{\tan }^{-1}}\frac{4}{3}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
E) \[{{\tan }^{-1}}\frac{3}{4}\]
Correct Answer: E
Solution :
Given curve are \[y={{x}^{2}}\] ...(i) and \[{{y}^{2}}-x=0\] ...(ii) From Eq. (i), \[\frac{dy}{dx}=2x\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}=2={{m}_{1}}\] (say) From Eq. (ii), \[2y\frac{dy}{dx}-1=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2y}\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}=\frac{1}{2}={{m}_{2}}\] (say) Let\[\theta \]be the angle between given curve. Then, \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\] \[=\frac{2-\frac{1}{2}}{1+2.\frac{1}{2}}=\frac{3/2}{2}=\frac{3}{4}\] \[\therefore \] \[\theta ={{\tan }^{-1}}\frac{3}{4}\]
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