Applications-Of-Derivatives Question 451
Question: The angle between curves $ y^{2}=4x $ and $ x^{2}+y^{2}=5 $ at (1, 2) is
[Karnataka CET 1999]
Options:
A) $ {{\tan }^{-1}}(3) $
B) $ {{\tan }^{-1}}(2) $
C) $ \frac{\pi }{2} $
D) $ \frac{\pi }{4} $
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Answer:
Correct Answer: A
Solution:
For curve $ y^{2}=4x $
Þ $ \frac{dy}{dx}=\frac{4}{2y} $ \ $ {{( \frac{dy}{dx} )}{(1,,2)}}=1 $ and for curve $ x^{2}+y^{2}=5 $
Þ $ \frac{dy}{dx}=\frac{-x}{y} $ \ $ {{( \frac{dy}{dx} )}{(1,,2)}}=\frac{-1}{2} $ \ Angle between the curves is $ \theta ={{\tan }^{-1}}| ,\frac{\frac{-1}{2},-1}{1+( \frac{-1}{2} )}, |={{\tan }^{-1}}(3) $ .
BETA
