xcostheta/a +ysintheta/b=1, ax/costheta-by/sintheta =a^2-b^2, then prove that x^2/a^2 +y^2/b^2 =1

Asked by tarunkumardas32 | 13th Aug, 2020, 09:13: AM

Expert Answer:

Given: xcostheta/a +ysintheta/b=1, ax/costheta-by/sintheta =a2-b2 
TO prove: x2/a2 + y2/b2 =1
xcosθ over straight a plus ysinθ over straight b equals 1 space rightwards double arrow space bxcosθ space plus space aysinθ space equals space ab space... space left parenthesis straight i right parenthesis
ax over cosθ minus by over sinθ equals straight a squared minus straight b squared space space rightwards double arrow space axsinθ space minus space bycosθ equals open parentheses straight a squared minus straight b squared close parentheses sinθcosθ space... space left parenthesis ii right parenthesis
Multiplying space left parenthesis straight i right parenthesis space by space space bcosθ space space and space left parenthesis ii right parenthesis space by space space asinθ space respectively comma space we space get
straight b squared xcos squared straight theta space plus space abysinθ space cosθ space equals space ab squared cosθ space... space left parenthesis iii right parenthesis
straight a squared xsin squared straight theta space minus space abycosθ space sinθ space equals space open parentheses straight a squared minus straight b squared close parentheses asin squared straight theta space cosθ space... space left parenthesis iv right parenthesis
Adding space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space get
straight b squared xcos squared straight theta plus straight a squared xsin squared straight theta space equals space ab squared cosθ space plus thin space open parentheses straight a squared minus straight b squared close parentheses asin squared straight theta space cosθ
rightwards double arrow straight b squared xcos squared straight theta plus straight a squared xsin squared straight theta space equals space ab squared cosθ space plus thin space straight a cubed sin squared straight theta space cosθ space minus space straight b squared asin squared straight theta space cosθ
rightwards double arrow straight x open parentheses straight b squared cos squared straight theta plus straight a squared sin squared straight theta close parentheses equals ab squared cosθ open parentheses 1 minus sin squared straight theta close parentheses plus thin space straight a cubed sin squared straight theta space cosθ space
rightwards double arrow straight x open parentheses straight b squared cos squared straight theta plus straight a squared sin squared straight theta close parentheses equals ab squared cos cubed straight theta plus thin space straight a cubed sin squared straight theta space cosθ
rightwards double arrow straight x open parentheses straight b squared cos squared straight theta plus straight a squared sin squared straight theta close parentheses equals acosθ open parentheses straight b squared cos squared straight theta plus thin space straight a squared sin squared straight theta close parentheses
rightwards double arrow straight x equals acosθ space... space left parenthesis straight v right parenthesis
Substituting space in space left parenthesis straight i right parenthesis comma space we space get
straight y equals bsinθ space... space left parenthesis vi right parenthesis
Consider comma
straight x squared over straight a squared plus straight y squared over straight b squared equals fraction numerator straight a squared cos squared straight theta over denominator straight a squared end fraction plus fraction numerator straight b squared sin squared straight theta over denominator straight b squared end fraction equals sin squared straight theta plus cos squared straight theta equals 1
Hence comma space straight x squared over straight a squared plus straight y squared over straight b squared equals 1

Answered by Renu Varma | 14th Aug, 2020, 12:01: PM

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