Applications Of Derivatives Question 63
Question: If OT is the perpendicular drawn from the origin to the tangent at any point t to the curve $ x=acos^{3}t,y=asin^{3}t $ , then OT is equal to:
Options:
A) a sin 2t
B) $ \frac{a}{2}\sin 2t $
C) 2a sin 2t
D) 2a
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3a{{\sin }^{2}}t\cos t}{-3a{{\cos }^{2}}t\sin t}=-\tan t $
$ \therefore $ equation of the tangent at -t- is $ y-a{{\sin }^{3}}t=-\tan t(x-a{{\cos }^{3}}t) $
$ \Rightarrow x\tan t+y-a({{\sin }^{3}}t+\sin ,t.{{\cos }^{2}}t)=0 $
$ \Rightarrow x\tan t+y-a\sin t=0 $
$ \therefore $ distance from the origin to this tangent $ =\frac{| -a\sin t |}{\sqrt{{{\tan }^{2}}t+1}}=\frac{a\sin t}{\sec t}=\frac{a}{2}\sin 2t $
BETA
