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Vector Algebra

Question
CBSEENMA12032918

Represent the following family of curves by forming the corresponding differential equations (a. b: parameters).
straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1



Solution

The equation of curve is
           straight x squared over straight a squared plus straight y squared over straight b squared space equals space 1                          ...(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 yy subscript 1 over denominator straight b squared end fraction equals 0   or     straight x over straight a squared plus fraction numerator straight y space straight y subscript 1 over denominator straight b squared end fraction equals 0           ...(2)
rightwards double arrow space space space space space space space space space space space straight a squared over straight b squared 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 space plus space straight y subscript 1 superscript 2 over denominator space 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 space space space space space space space space space space space rightwards double arrow 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 straight x space left parenthesis straight y space straight y subscript 2 plus space 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.