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Home / Doubts and Solutions/JEE/Mathematics

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Asked by rbhatt16 22nd June 2018, 12:11 AM
Answered by Expert
Answer:
begin mathsize 16px style xcosα plus ysinα equals straight p.......... left parenthesis straight i right parenthesis ysinα equals straight p minus xcosα straight y equals fraction numerator straight p minus xcosα over denominator sin straight alpha end fraction equals pcosec straight alpha minus xcotα Line space left parenthesis straight i right parenthesis space is space tangent space to space the space curve straight x squared over straight a squared plus straight y squared over straight b squared equals 1........... left parenthesis ii right parenthesis straight b squared straight x squared plus straight a squared straight y squared equals straight a squared straight b squared Put space straight y equals pcosecα minus xcotα space in space the space above space equation space and space simplify straight x squared left parenthesis straight b squared plus straight a squared cot squared straight alpha right parenthesis minus 2 straight a squared xpcosecαcotα plus straight a squared left parenthesis straight p squared cosec squared straight alpha minus straight b squared right parenthesis equals 0 For space tangent space it space must space have space equal space roots. straight D equals 0 Simplifying space we space get space straight a squared cos squared straight alpha plus straight b squared sin squared straight alpha equals straight p squared end style
Answered by Expert 25th June 2018, 11:27 AM
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