A) 2
B) \[\sqrt{2}\]
C) \[2\sqrt{2}\]
D) 4
Correct Answer: A
Solution :
Given that \[x=2\cos t+2t\sin t\] So, \[\frac{dx}{dt}=-2\sin t+2\left[ t\cos t+\sin t \right]\] \[\frac{dx}{dt}=2\cos t-2\left[ -t\sin t+\cos t \right]\] \[\frac{dx}{dt}=2t\sin t\] ?.(ii) \[\frac{dy}{dx}=\frac{2t\sin t}{2t\sin t}\] \[\frac{dy}{dx}=\tan t\] \[{{\left( \frac{dy}{dx} \right)}_{t=1/4}}=1\] so the slopw of the normal is -1 and \[atl=\pi /4x=\sqrt{2}+\frac{\pi }{2\sqrt{2}}\]and\[y=\sqrt{2}-\pi /2\sqrt{2}\]to the equation of normal is\[\left[ y-\left( \sqrt{2}-\pi /2\sqrt{2} \right) \right]=-1\left[ \left( x-\left( \sqrt{2}+\pi /2\sqrt{2} \right) \right) \right]\]\[y-\sqrt{2}+\frac{\pi }{2\sqrt{2}}=-x+\sqrt{2}+\pi /2\sqrt{2}\] \[x+y=2\sqrt{2}\] so the distacne from the origin is 2
You need to login to perform this action.
You will be redirected in
3 sec
