← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Orthogonal trajectories (cartesian and polar) · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Show that the family of parabolas y2=4cx+4c2y^2=4cx+4c^2 is self-orthogonal.

Technique

Eliminate cc to get the family ODE; substitute y′→−1/y′y'\to-1/y'; show the ODE returns (self-orthogonality).

Solution

A one-parameter family is self-orthogonal if its differential equation is unchanged when the slope y′y' is replaced by −1/y′-1/y' (the substitution that converts a curve into its orthogonal trajectory). We first form the ODE of the family by eliminating cc.

Step 1 — Form the differential equation of the family

y2=4cx+4c2.(1)y^2=4cx+4c^2.\tag{1}

Differentiate w.r.t. xx (write p=y′p=y'):

2yp=4c⟹c=yp2.(2)2yp=4c\quad\Longrightarrow\quad c=\frac{yp}{2}.\tag{2}

Step 2 — Eliminate cc

Substitute (2) into (1):

y2=4⋅yp2 x+4(yp2)2=2xyp+y2p2.y^2=4\cdot\frac{yp}{2}\,x+4\left(\frac{yp}{2}\right)^2=2xyp+y^2p^2.

Hence the differential equation of the family is

y2p2+2xyp−y2=0,i.e.y2 y′2+2xy y′−y2=0.(3)y^2 p^2+2xyp-y^2=0,\qquad\text{i.e.}\qquad y^2\,y'^2+2xy\,y'-y^2=0.\tag{3}

(Dividing by y2y^2:   p2+2xyp−1=0\;p^2+\dfrac{2x}{y}p-1=0.)

Step 3 — Replace y′→−1y′y'\to-\dfrac{1}{y'}

For the orthogonal trajectories, substitute p→−1pp\to-\dfrac1p in (3):

y2(−1p)2+2xy(−1p)−y2=0.y^2\left(-\frac1p\right)^2+2xy\left(-\frac1p\right)-y^2=0.

Multiply through by p2p^2:

y2−2xyp−y2p2=0⟹y2p2+2xyp−y2=0.y^2-2xyp-y^2p^2=0\quad\Longrightarrow\quad y^2p^2+2xyp-y^2=0.

Step 4 — Compare

This is identical to equation (3). The differential equation of the orthogonal trajectories is the same as that of the original family, so the family of parabolas is its own set of orthogonal trajectories:

Answer

  y2 y′2+2xy y′−y2=0  is invariant under y′↦−1/y′ ⇒ the family is self-orthogonal.  \boxed{\;y^2\,y'^2+2xy\,y'-y^2=0\ \text{ is invariant under } y'\mapsto-1/y'\ \Rightarrow\ \text{the family is self-orthogonal.}\;}
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