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Vector Algebra

Question
CBSEENMA12032931

Form the differential equation corresponding to y2 – 2 a y + x2 – a2 by eliminating a. 

Solution
The given equation is
                                  straight y squared minus 2 ay plus straight x squared space equals space straight a squared               ...(1)
Differentiating both sides w.r.t.x, we get, 
                  2 space straight y space dy over dx minus 2 space straight a dy over dx plus 2 straight x equals 0 space space space or space space straight y dy over dx plus straight x space equals space straight a dy over dx
therefore space space space straight a space equals space fraction numerator straight y begin display style dy over dx end style plus straight x over denominator begin display style dy over dx end style end fraction                                                 ...(2)
Eliminating m from (1) and (2), we get,
     straight y squared minus fraction numerator 2 straight y squared begin display style dy over dx end style plus 2 xy over denominator begin display style dy over dx end style end fraction plus straight x squared space equals space open parentheses straight y begin display style dy over dx end style plus straight x close parentheses squared over open parentheses begin display style dy over dx end style close parentheses squared
or        open parentheses straight x squared plus straight y squared close parentheses open parentheses dy over dx close parentheses squared minus 2 straight y squared open parentheses dy over dx close parentheses squared minus 2 xy open parentheses dy over dx close parentheses space equals space open parentheses straight y dy over dx plus straight x close parentheses squared
or          open parentheses straight x squared plus straight y squared close parentheses open parentheses dy over dx close parentheses squared minus 2 straight x squared open parentheses dy over dx close parentheses squared minus 2 xy dy over dx equals straight y squared open parentheses dy over dx close parentheses squared plus straight x squared plus 2 xy dy over dx
or           open parentheses straight x squared minus 2 straight y squared close parentheses space open parentheses dy over dx close parentheses squared space minus space 4 xy dy over dx minus straight x squared space equals space 0
or        left parenthesis 2 straight y squared minus straight x squared right parenthesis space open parentheses dy over dx close parentheses squared plus 4 xy dy over dx plus straight x squared space equals space 0
which is required differential equation. 

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