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Vector Algebra

Question
CBSEENMA12032982

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

Solution
The general equation of such ellipse is
straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1 space space left parenthesis straight a comma space space straight b space being space arbitrary space constants right parenthesis     ...(1)

Differentiating w.r.t. x successively two times, we get.
               fraction numerator 2 straight x over denominator straight a squared end fraction plus fraction numerator 2 straight y space straight y subscript 1 over denominator straight b squared end fraction space equals space 0 space space space space or space space straight x over straight a squared plus fraction numerator straight y space straight y subscript 1 over denominator straight b squared end fraction space equals space 0              ...(2)
rightwards double arrow space space space space straight a squared over straight b squared space equals space minus fraction numerator straight x over denominator straight y space straight y subscript 1 end fraction                                                               ...(3)
and 1 over straight a squared plus fraction numerator straight y space straight y subscript 2 plus space straight y subscript 1 squared over denominator straight b squared end fraction space equals space 0 space space space space space space space space rightwards double arrow space space space space straight a squared over straight b squared space equals space fraction numerator negative 1 over denominator straight y space straight y subscript 2 space plus space straight y subscript 1 squared end fraction             ...(4)
From (3), (4), we get
                     fraction numerator negative straight x over denominator straight y space straight y subscript 1 end fraction space equals space fraction numerator negative 1 over denominator straight y space straight y subscript 2 space plus space straight y subscript 1 squared end fraction space space space space space space rightwards double arrow space space space space space straight x left parenthesis straight y space straight y subscript 2 plus straight y subscript 1 squared right parenthesis space minus space straight y space straight y subscript 1 space equals space 0
which is the required differential equation.