Conic Sections Question 69
Question: The angle of intersection between the curves $ y^{2}=4x $ and $ x^{2}=32y $ at point (16, 8), is
[RPET 1987, 96]
Options:
A) $ {{\tan }^{-1}}( \frac{3}{5} ) $
B) $ {{\tan }^{-1}}( \frac{4}{5} ) $
C) $ \pi $
D) $ \frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Tangent at (16,8) to both are $ 8y=2(x+16) $
….. (i) and $ 16x=16(y+8) $ …..(ii) $ m_1=\frac{1}{4},m_2=1 $
$ \tan \theta =\frac{m_2-m_1}{1+m_2m_1}=( \frac{3}{5} ) $
therefore $ \theta ={{\tan }^{-1}}( \frac{3}{5} ) $ . Aliter: Using direct formula $ \theta ={{\tan }^{-1}}\frac{3{a^{1/3}}{b^{1/3}}}{2({a^{2/3}}+{b^{2/3}})}, $ where $ a=1 $ and $ b=8 $
$ ={{\tan }^{-1}}\frac{6}{2(1+4)}={{\tan }^{-1}}\frac{3}{5} $ .
BETA
