Question

Find the orthogonal trajectories of the family of curves \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\) where k is the parameter.

(a) x^2-y^2-3a^2  log⁡x-k = 0

(b) x^2+2y^2–\(\frac{a^2}{2}\)  log⁡x-k = 0

(c) x^2+y^2-2a^2  log⁡x-k = 0

(d) 2x^2-y^2–\(\frac{a^2}{3}\)  log⁡x-k = 0

The question was asked in an online quiz.

My question is based upon Orthogonal Trajectories topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (c) x^2+y^2-2a^2  log⁡x-k = 0

For explanation I would say: \(\frac{x^2}{a^2} + \frac{y^2}{b^2+k} = 1\)…….(1) differentiating w.r.t x we have \(\frac{2x}{a^2} + \frac{2yy’}{b^2+k} = 0 \,where\, y’ = \frac{dy}{dx}\)

 i.e \(\frac{x^2}{a^2} = \frac{-yy’}{b^2+k} …..(2) \,from\, (1)\, \frac{x^2}{a^2} – 1 = \frac{-y^2}{b^2+k} \rightarrow \frac{x^2-a^2}{a^2} = \frac{-y^2}{b^2+k}\)…..(3) divide (2)&(3)

we get \(\frac{x}{x^2-a^2} = \frac{yy’}{y^2} \rightarrow \frac{x}{x^2-a^2} = \frac{y’}{y}\)  now \(y’=\frac{dy}{dx}\) is replaced by \(-\frac{dx}{dy} \,i.e\, \frac{x}{x^2-a^2} = \frac{1}{y}(-\frac{dx}{dy})\)

separating the variable we get  \(y \,dy = -\frac{(x^2-a^2)}{x} dx\) integrating this equation

\(\int y \,dy = \int-x \,dx + \int a^2 \frac{1}{x} \,dx + c \rightarrow \frac{y^2}{2} = \frac{-x^2}{2} + a^2  log⁡ \,x + c\)

–> x^2+y^2-2a^2 log⁡x – 2c=0 or x^2+y^2-2a^2 log⁡x-k=0 where k=2c is the  required orthogonal trajectory.