Applications-Of-Derivatives Question 458
Question: The normal of the curve $ x=a(\cos \theta +\theta \sin \theta ) $ $ y=a(\sin \theta -\theta \cos \theta ) $ at any $ \theta $ is such that
[DCE 2000; AIEEE 2005]
Options:
A) It makes a constant angle with x-axis
B) It passes through the origin
C) It is at a constant distance from the origin
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=a(\sin \theta -\theta \cos \theta ),,x=a(\cos \theta +\theta \sin \theta ) $ $ \frac{dy}{d\theta }=a[\cos \theta -\cos \theta +\theta \sin \theta ]=a\theta \sin \theta $ $ \frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )=a\theta \cos \theta $
$ \therefore ,\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta $
Þ Slope of the tangent = $ \tan \theta $ \ Slope of the normal = $ -\cot \theta $ Hence, equation of normal $ [y-a\sin \theta +a\theta \cos \theta ]=-\frac{\cos \theta }{\sin \theta } $ $ [x-a\cos \theta -a\theta \sin \theta ] $
Þ $ y\sin \theta -a{{\sin }^{2}}\theta +a\theta \sin \theta \cos \theta $ $ =-x\cos \theta +a{{\cos }^{2}}\theta +a\theta \sin \theta \cos \theta $
Þ $ x\cos \theta +y\sin \theta =a({{\sin }^{2}}\theta +{{\cos }^{2}}\theta ) $
Þ $ x\cos \theta +y\sin \theta =a $ \Distance from origin = $ \frac{a}{\sqrt{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }}=\text{a constant} $
BETA
