Question

Find the orthogonal trajectories of the family of parabolas y^2=4ax.

(a) 2x^2+y^2=k

(b) 2y^2+x^2=k

(c) x^2-2y^2=k

(d) 2x^2-y^2=k

The question was asked during an online interview.

This intriguing question comes from Orthogonal Trajectories in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (a) 2x^2+y^2=k

To elaborate: Consider \(\frac{y^2}{x} = 4a\)……..differentiating w.r.t x we get \(\frac{x.2y \frac{dy}{dx} – y^2.1}{x^2} = 0\)

\(2xy \frac{dy}{dx} -y = 0\)…..DE of the family of curve y^2=4ax replacing \(\frac{dy}{dx}\)  by \(-\frac{dx}{dy}\)

since \(\frac{dy}{dx}\)  is the slope of the tangent to the curve –> \(-\frac{dx}{dy}\) is the slope of orthogonal line we get \(2x(-\frac{dx}{dy}) – y = 0 \, or\, 2x \,dx + y \,dy = 0 \rightarrow \int2x \,dx + \int y \,dy = c \rightarrow x^2 + \frac{y^2}{2} = c.\)

or 2x^2+y^2=k is the required orthogonal trajectory.