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Vector Algebra

Question
CBSEENMA12032985

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

Solution
The equation of family of hyperbolas having foci on x-axis and centre at origin is
              straight x squared over straight a squared minus straight y squared over straight b squared space equals 1                             ...(1)
Differentiating w.r.t.. x successively two times, we get,
                           fraction numerator 2 straight x over denominator straight a squared end fraction minus fraction numerator 2 space straight y space straight y subscript 1 over denominator straight b squared end fraction space equals space 0 space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space straight x over straight a squared minus fraction numerator straight y space straight y subscript 1 over denominator straight b squared end fraction space equals space 0
rightwards double arrow 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 fraction numerator straight x over denominator straight y space straight y subscript 1 end fraction                ...(2)
and     1 over straight a squared minus fraction numerator straight y space straight y subscript 2 plus straight y subscript 1 squared over denominator straight b squared end fraction space equals space 0 space space space space space space space 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 1 over denominator straight y space straight y subscript 2 space plus space straight y subscript 1 squared end fraction              ...(3)
From (2), (3), we get,
                         fraction numerator straight x over denominator straight y space straight y subscript 1 end fraction space equals space fraction numerator 1 over denominator straight y space straight y subscript 2 plus straight y subscript 1 squared end fraction space space or space space straight x thin space 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.