Search any question & find its solution
Question:
Answered & Verified by Expert
The normal to the curve $x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)$ at any point ' $\theta$ ' is such that
Options:
Solution:
1410 Upvotes
Verified Answer
The correct answer is:
it is at a constant distance from the origin
it is at a constant distance from the origin
Clearly $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta \Rightarrow$ slope of normal $=-\cot \theta$
Equation of normal at ' $\theta$ ' is
$y-a(\sin \theta-\theta \cos \theta)=-\cot \theta(x-a(\cos \theta+\theta \sin \theta)$
$\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \cos \theta \sin \theta=-x \cos \theta+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$
$\Rightarrow x \cos \theta+y \sin \theta=a$
Clearly this is an equation of straight line which is at a constant distance 'a' from origin.
Equation of normal at ' $\theta$ ' is
$y-a(\sin \theta-\theta \cos \theta)=-\cot \theta(x-a(\cos \theta+\theta \sin \theta)$
$\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \cos \theta \sin \theta=-x \cos \theta+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$
$\Rightarrow x \cos \theta+y \sin \theta=a$
Clearly this is an equation of straight line which is at a constant distance 'a' from origin.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.