JEE Mains · Maths · STD 12 - 6. Application of derivatives
The distance, from the origin, of the normal to the curve, \(x = 2\,cos\,t + 2t\,sin\,t,\,\,y = 2\,sin\,t -2t\, cos\,t\) at \(t= \frac {\pi }{4},\) is
- A \(2\)
- B \(4\)
- C \(\sqrt 2\)
- D \(2\sqrt 2\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
Given that \(x = 2\cos t + 2t\sin t\) so, \(\frac{{dx}}{{dt}} = - 2\sin t + 2\left[ {t\cos t + \sin t} \right]\) \(\frac{{dy}}{{dt}} = 2\cos t - 2\left[ { - t\sin t + \cos t} \right]\) \(\frac{{dy}}{{dt}} = 2t\sin t\,\,\,\,\,\,\,\,....\left( {ii} \right)\)…
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