Conic Sections Question 481
Question: If q is the acute angle of intersection at a real point of intersection of the circle $ x^{2}+y^{2}=5 $ and the parabola $ y^{2}=4x $ then tanq is equal to
[Karnataka CET 2005]
Options:
A) 1
B) $ \sqrt{3} $
C) 3
D) $ \frac{1}{\sqrt{3}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Solving equations $ x^{2}+y^{2}=5 $ and $ y^{2}=4x $
we get $ x^{2}+4x-5=0 $ i.e., $ x=1,-5 $
For $ x=1 $ ; $ y^{2}=4 $
therefore $ y=\pm 2 $
For $ x=-5 $ ; $ y^{2}=-20 $ (imaginary values) Points are (1, 2)(1, -2); $ m_1 $ for $ x^{2}+y^{2}=5 $ at (1, 2) $ {{. \frac{dy}{dx}=-\frac{x}{y} |} _{(1,2)}}=-\frac{1}{2} $ Similarly, $ m_2 $ for $ y^{2}=4x $ at (1,2) is 1. $ \tan \theta =| \frac{m_1-m_2}{1+m_1m_2} |=| \frac{-\frac{1}{2}-1}{1-\frac{1}{2}} |=3 $ .
BETA
