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Vector Algebra

Question
CBSEENMA12032984

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Solution
The equation of family of ellipses having foci on y-axis and centre at origin is
                    straight x squared over straight a squared plus straight y squared over straight a squared space equals space 1                            ...(1)
 Differentiating w.r.t. x, we get,
                            fraction numerator 2 straight x over denominator straight b squared end fraction plus fraction numerator 2 straight y over denominator straight a squared end fraction dy over dx space equals space 0              ...(2)
therefore space space space space space space space straight a squared over straight b squared space equals space fraction numerator negative straight y over denominator straight x end fraction dy over dx                                                              ...(3)
Again differentiating (2) w.r.t.x, we get,
                         2 over straight b squared plus fraction numerator 2 straight y over denominator straight a squared end fraction fraction numerator straight d squared straight y over denominator dx squared end fraction plus 2 over straight a squared open parentheses dy over dx close parentheses squared space space equals space 0
therefore space space space space space space space space space space space space space space space space space space space space space space space space space space space straight a squared over straight b squared space equals space minus open square brackets straight y fraction numerator straight d squared straight y over denominator dx squared end fraction plus open parentheses dy over dx close parentheses squared close square brackets                       ...(4)
From (3) and (4), we get,
   space space space space space space space space space space space space space space space space space space space space space space space space minus straight y over straight x dy over dx space equals space minus open square brackets straight y fraction numerator straight d squared straight y over denominator dx squared end fraction plus open parentheses dy over dx close parentheses squared close square brackets
therefore space space space space space space space space space space space space space space space space space space space space straight y over straight x dy over dx equals space straight x space straight y space fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x open parentheses dy over dx close parentheses squared
which is required differential equation.