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UPSC 2022 Maths Optional Paper 1 Q5b — 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 orthogonal trajectories of the system of parabolas $x^2=4a(y+a)$ belong to the same system.

Technique

Differentiate-and-eliminate to get DE of family; substitute $y'\to-1/y'$ for orthogonal family; observe the resulting DE is the same.

Solution

Setup. Differentiate to find $dy/dx$ of the family, eliminate $a$, then replace $dy/dx\to-dx/dy$ (orthogonal trajectory) and solve.

Step 1 — Differentiate $x^2=4a(y+a)$

$2x=4a y'$, so $a=\dfrac{x}{2y'}$.

Substitute back: $x^2=4\cdot\dfrac{x}{2y'}\cdot\left(y+\dfrac{x}{2y'}\right)=\dfrac{2x}{y'}\!\left(y+\dfrac{x}{2y'}\right)=\dfrac{2xy}{y'}+\dfrac{x^2}{y'^2}$.

Multiply by $y'^2$:

Divide by $x^2$ (assuming $x\ne 0$):

Step 2 — Replace $y'\to -1/y'$ for orthogonal trajectories

Let $y'_{\perp}=-1/y'$ (orthogonal slope). Substituting into $(\star)$:

$\left(-\dfrac{1}{y'}\right)^2-\dfrac{2y}{x}\!\left(-\dfrac{1}{y'}\right)-1=0$,

$\dfrac{1}{y'^2}+\dfrac{2y}{x y'}-1=0$.

Multiply by $y'^2$:

Hmm — this is exactly the same equation as $(\star)$:

$y'^2-\dfrac{2y}{x}y'-1=0$ (the original equation rearranged).

Wait — let me re-examine. The original was $y'^2-(2y/x)y'-1=0$. The “orthogonal” equation came out to $y'^2-(2yy'/x)-1=0$.

These look the same? Let me reread: $\dfrac{2y}{x}y'$ vs $\dfrac{2yy'}{x}$ — these are identical.

So $(\star)$ is invariant under $y'\to-1/y'$ (after multiplying by $y'^2$ and substituting).

Conclusion. The differential equation for the orthogonal trajectories is the same as the original DE. Hence the orthogonal trajectories solve the same DE, which gives the same family of parabolas $x^2=4a(y+a)$.

Answer