Question

The Orthogonal DE for family of  parabola y^2=4a(x+a) is same as _______(where DE stands for Differential equation)

(a) DE of parabola y^2=4a(x+a)

(b) DE of parabola y^2=4ax

(c) DE of parabola x^2=4ay

(d) DE of parabola x^2=4a(y+a)

I got this question in final exam.

The above asked question is from Orthogonal Trajectories in chapter Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right choice is (a) DE of parabola y^2=4a(x+a)

Explanation: y^2=4a(x+a)……….differentiating w.r.t x we get \(2y \frac{dy}{dx} = 4a \rightarrow a =\frac{yy’}{2}\)

 substituting the ‘a’ in given equation i.e : \(y^2 = 2yy’(x + \frac{yy’}{2}), y=2xy’+yy’^2\)…(1)

replacing y’ by \(\frac{-1}{y’} \,i.e\, y=2x\frac{-1}{y’} + y(\frac{-1}{y’})^2\) or on solving yy’^2+2xy’=y…(2)

comparing (2) & (1) we observe that DE of the orthogonal family is same as the DE of the given family of curves thus the family of parabola is self orthogonal.