-->

Vector Algebra

Question
CBSEENMA12032920

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




Solution
The general equation of hyperbolas is
straight x squared over straight a squared minus straight y squared over straight b squared 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 space space space space space space rightwards double arrow 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 straight a squared over straight b squared space equals space 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 space 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 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 space straight y subscript 1 squared end fraction space space space space or space space space straight x space left parenthesis straight y space straight y subscript 2 space 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.

Some More Questions From Vector Algebra Chapter