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Vector Algebra

Question
CBSEENMA12032943

Form the differential equation of the family of curves represented by the equation (x – a)2 + 2 y2 = a2, a being the parameter.

Solution
The given equation is
                  open parentheses straight x minus straight a close parentheses squared plus 2 straight y squared space equals space straight a squared                    ...(1)
Differentiating both sides w.r.t.x , we get.
                     2 left parenthesis straight x minus straight a right parenthesis plus 4 straight y dy over dx space equals 0 
therefore space space                              straight x minus straight a equals negative 2 straight y dy over dx                 ...(2)
and                             straight a equals straight x plus 2 straight y dy over dx                       ...(3)
Putting the values of x - a, a from (2), (3) in (1), we get, 
                     open parentheses negative 2 straight y space dy over dx close parentheses squared plus 2 straight y squared space equals space open parentheses straight x plus 2 straight y dy over dx close parentheses squared
or            4 straight y squared open parentheses dy over dx close parentheses squared plus 2 straight y squared space equals space straight x squared plus 4 straight y squared open parentheses dy over dx close parentheses squared plus 4 xy dy over dx
or                                         2 straight y squared space equals space straight x squared plus 4 xy dy over dx
or                               2 straight y squared minus straight x squared space equals space 4 xy dy over dx
which is the required differential equation of the given family of curves.