A) \[{{\tan }^{-1}}(3/4)\]
B) \[{{\tan }^{-1}}(4/5)\]
C) \[{{\tan }^{-1}}(2/3)\]
D) none of these
Correct Answer: A
Solution :
Equation of curves \[x={{y}^{2}}\]and \[y={{x}^{2}}\] \[2y\frac{dy}{dx}=1\Rightarrow \frac{dy}{dx}=\frac{1}{2y}=\frac{1}{2}\] \[\frac{dy}{dx}=2x=2\] \[{{m}_{2}}=1/2\] \[{{m}_{1}}=2\] \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}} \right|=\left| \frac{3}{4} \right|\] \[\theta ={{\tan }^{-1}}3/4\]
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