Straight Line Question 392
Question: The line which is parallel to x?axis and crosses the curve $ y=\sqrt{x} $ at an angle of $ 45^{o} $ is equal to [Pb. CET 2002]
Options:
A) $ x=\frac{1}{4} $
B) $ y=\frac{1}{4} $
C) $ y=\frac{1}{2} $
D) $ y=1 $
Correct Answer: C $ \therefore $ (i) and (ii) intersect at P, at $ 45{}^\circ $ $ \therefore $ $ {{\tan }^{-1}}( \frac{m-0}{1+m.0} )=\pm 45{}^\circ $ . $ \therefore $ The equation of line is $ y=\frac{1}{2} $ or $ y=\frac{-1}{2} $ but $ y=\frac{-1}{2} $ is not given, hence the required line is $ y=\frac{1}{2} $ .Show Answer
Answer:
Solution:
$ \therefore $ Slope of (ii) is, m= $ {{( \frac{dy}{dx} )} _{\text{at }P}}=\frac{1}{2\lambda } $
Þ $ m=( \frac{1}{2\lambda } )=\pm 1 $
Þ $ \lambda =\pm \frac{1}{2} $
BETA
