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CBSE Class 11-science Answered

14th sum plz

Asked by lovemaan5500 | 07 Oct, 2018, 02:35: PM
Expert Answer
begin mathsize 16px style straight x over straight a cosθ 6 plus straight y over straight b sinθ equals 1
bxcosθ plus aysinθ equals ab
bxcosθ plus aysinθ minus ab equals 0...... left parenthesis straight i right parenthesis
Length space of space tehperpendicular space from space the space point space open parentheses square root of straight a squared minus straight b squared end root comma 0 close parentheses space is
straight p subscript 1 space equals space fraction numerator open vertical bar bcosθ left parenthesis square root of straight a squared minus straight b squared end root plus asinθ cross times 0 minus ab close vertical bar over denominator square root of straight b squared cos squared straight theta plus straight a squared sin squared straight theta end root end fraction
straight p subscript 1 equals fraction numerator begin display style open vertical bar bcosθ left parenthesis square root of straight a squared minus straight b squared end root minus ab close vertical bar end style over denominator begin display style square root of straight b squared cos squared straight theta plus straight a squared sin squared straight theta end root end style end fraction........ left parenthesis ii right parenthesis
Similarly space find space straight p subscript 2 space end subscript equals space fraction numerator begin display style open vertical bar bcosθ left parenthesis square root of straight a squared minus straight b squared end root plus ab close vertical bar end style over denominator begin display style square root of straight b squared cos squared straight theta plus straight a squared sin squared straight theta end root end style end fraction
find space straight p subscript 1 straight p subscript 2 space we space get space
straight p subscript 1 straight p subscript 2 space equals fraction numerator begin display style open vertical bar bcosθ left parenthesis square root of straight a squared minus straight b squared end root plus asinθ cross times 0 minus ab close vertical bar end style over denominator begin display style square root of straight b squared cos squared straight theta plus straight a squared sin squared straight theta end root end style end fraction cross times space fraction numerator begin display style open vertical bar bcosθ left parenthesis square root of straight a squared minus straight b squared end root plus ab close vertical bar end style over denominator begin display style square root of straight b squared cos squared straight theta plus straight a squared sin squared straight theta end root end style end fraction
Simplifying space this space we space get space
straight p subscript 1 straight p subscript 2 space equals straight b squared end style
Answered by Sneha shidid | 08 Oct, 2018, 11:37: AM
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