Applications Of Derivatives Question 132
Question: The angle of intersection of the curves $ y=x^{2} $ and $ x=y^{2} $ at (1, 1) is
[Roorkee 2000; Karnataka CET 2001]
Options:
A) $ {{\tan }^{-1}}( \frac{4}{3} ) $
B) $ {{\tan }^{-1}}(1) $
C) $ 90^{o} $
D) $ {{\tan }^{-1}}( \frac{3}{4} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=x^{2} $
therefore $ \frac{dy}{dx}=m_1=2x $
therefore $ {{( \frac{dy}{dx} )} _{(1,,1)}}=2=m_1 $ and $ x=y^{2} $
therefore $ 1=2y,\frac{dy}{dx} $
therefore $ \frac{dy}{dx}=m_2=\frac{1}{2y} $
therefore $ {{( \frac{dy}{dx} )} _{(1,,1)}}=\frac{1}{2} $
$ \therefore $ Angle of intersection, $ \tan \theta =\frac{m_1-m_2}{1+m_1m_2} $ = $ \frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}} $ = $ \frac{3}{4} $
therefore $ \theta ={{\tan }^{-1}}(3/4) $ .
BETA
