Conic Sections Question 430
Question: The angle of intersection of the curves $ y^{2}=2x/\pi $ and $ y=\sin x $ , is
[Roorkee 1998]
Options:
A) $ {{\cot }^{-1}}(-1/\pi ) $
B) $ {{\cot }^{-1}}\pi $
C) $ {{\cot }^{-1}}(-\pi ) $
D) $ {{\cot }^{-1}}(1/\pi ) $
Show Answer
Answer:
Correct Answer: B
Solution:
The curves $ y^{2}=2x/\pi $ and $ y=\sin x $ intersect at $ (0,0) $ and $ (\pi /2,1) $ . Let the gradients of the tangents to the curves be $ m_1 $ and $ m_2 $ respectively. Then $ m_1=\frac{dy}{dx}=\frac{1}{\pi y} $ and $ m_2=\frac{dy}{dx}=\cos x $
At $ (\pi /2,1) $ , $ m_1=\frac{1}{\pi } $ , $ m_2=\cos \frac{\pi }{2}=0 $
Thus $ \tan \theta =\frac{(1/\pi )-0}{1+(1/\pi )(0)}=\frac{1}{\pi } $
therefore $ \theta ={{\cot }^{-1}}\pi $ .
BETA
