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Vector Algebra

Question
CBSEENMA12032922

Represent the following family of curves by forming the corresponding differential equations (a. b: parameters).
straight y squared minus 4 straight a left parenthesis straight x minus straight b right parenthesis





Solution

The equation of given curve is
y2 = 4 a (x – b)
Differentiating w.r.t. x, we get,
2 straight y dy over dx space equals space 4 space straight a                or   straight y dy over dx space equals 2 straight a
Again differentiating w.r.t x, we get.
straight y fraction numerator straight d squared straight y over denominator dx squared end fraction plus dy over dx. dy over dx space equals space 0       or   straight y fraction numerator straight d squared straight y over denominator dx squared end fraction plus open parentheses dy over dx close parentheses squared space equals space 0
which is required differential equation..