The slope of the normal to the curve x = 1 - a sin , y = b cos2 θ ï»¿ï»¿ï»¿ï»¿ï»¿ at q = is

CBSE Class 12-science Answered

The slope of the normal to the curve x = 1 - a sin , y = b cos² θ ï»¿ï»¿ï»¿ï»¿ï»¿ at q = $straight pi over 2$ is

Asked by Topperlearning User | 07 Aug, 2014, 09:19: AM

Expert Answer

We have,

$straight x equals 1 minus straight a space sin space straight theta space & space straight y equals straight b space cos squared straight theta & space dx over dθ equals minus straight a space cos space straight theta space & space dy over dθ equals minus 2 straight b space cos space straight theta space sin space straight theta therefore dy over dx equals fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction equals fraction numerator minus 2 bsin space straight theta space cos space straight theta over denominator minus straight a space cos space straight theta end fraction equals fraction numerator 2 straight b over denominator straight a end fraction sin space straight theta rightwards double arrow open parentheses dy over dx close parentheses subscript straight theta equals straight pi over 2 end subscript equals fraction numerator 2 straight b over denominator straight a end fraction sin straight pi over 2 equals fraction numerator 2 straight b over denominator straight a end fraction Hence comma space open parentheses Slope space of space the space normal space at space straight theta equals straight pi over 2 close parentheses equals minus 1 over open parentheses begin display style dy over dx end style close parentheses subscript straight theta equals begin display style straight pi over 2 end style end subscript equals minus fraction numerator straight a over denominator 2 straight b end fraction$