Orthogonal Trajectories
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Direct answer
Families of curves that intersect another family everywhere at right angles are its orthogonal trajectories, and finding them is a two-step differential-equation routine: first extract the differential equation of the given family by differentiating its equation and eliminating the arbitrary constant; then replace dy/dx by −dx/dy (the negative reciprocal slope) in that differential equation and solve. Where the original family has slope m at a point, the orthogonal family must have slope −1/m, which is precisely the substitution. Thus the parabolas y = cx² have differential equation dy/dx = 2y/x, their orthogonal trajectories satisfy dy/dx = −x/(2y), and integrating gives x² + 2y² = C — ellipses cutting every parabola at right angles.
What you must remember
- The core substitution: slope m becomes −1/m, i.e. dy/dx → −dx/dy in the family's differential equation; this is the entire method in one symbol move.
- Eliminate the constant first: the given family y = f(x, c) must be reduced to a c-free differential equation before substituting — trajectories depend on the family, not on one member.
- Standard pairings: the orthogonal trajectories of the circles x² + y² = c are the straight lines y = kx through the origin (radial versus circular); of the family xy = c, the confocal hyperbolas x² − y² = k.
- Exponential family: trajectories of y = ce^x satisfy y' = y; the orthogonal equation y' = −1/y integrates to y² = −2x + C — sideways parabolas.
- Self-orthogonal caution: a family can be orthogonal to itself only in degenerate senses; if solving reproduces the original family, the constant elimination went wrong.
- Physical framing: in electrostatics and heat flow, equipotential curves and field lines (or isotherms and heat-flow lines) are mutually orthogonal trajectories — the standard application context.
- Verification habit: at a sample intersection point, the product of the two slopes must equal −1; a ten-second check that catches sign slips.
Finding the trajectory of y = cx²
Work the full pipeline. The family is y = cx² with parameter c. Differentiate: y' = 2cx. Eliminate c: from the original equation c = y/x², so y' = 2y/x — the differential equation of the family, now parameter-free. Substitute the negative reciprocal: y' = −x/(2y). Solve: 2y dy = −x dx, integrate to y² = −x²/2 + C', i.e. x² + 2y² = C. Verify at the point (2, 4): the parabola y = x² (c = 1) has slope 2x = 4; the ellipse through (2, 4) is x² + 2y² = 36 with slope from 2x + 4y·y' = 0 giving y' = −x/(2y) = −1/4. Product: 4 × (−1/4) = −1 — perpendicular confirmed.
The routine pays off exactly when each step is automatic: differentiate, eliminate, invert, integrate, verify. Note how the elimination step is the only place judgment enters — some families need algebraic work to remove c (solve for c from the original, substitute into the derivative). If the family is given already in differential form, skip straight to the inversion. An exam variation asks the question in reverse: which family has the ellipses x² + 2y² = C as orthogonal trajectories? Run the same pipeline and the parabolas re-emerge, since the relation is symmetric.
A niche that still appears
Orthogonal trajectories sit at the junction of differential equations and coordinate geometry, and JEE Advanced revisits them intermittently — directly as a five-mark differential-equations item, or disguised inside physics-flavoured questions where curves of equal potential meet curves of force. The classic error is substituting −1/y' into the family equation instead of into its differential equation: the negative reciprocal belongs to the slope field, not to the curve's algebraic equation. The second error is forgetting the constant of integration at the final step — the trajectory is a family, never a single curve, and an answer without + C is incomplete. The third, specific to implicit families, is differentiating without the chain rule: for x² + y² = c, differentiating gives 2x + 2y·y' = 0, and dropping the y factor on y' wrecks the inversion from the start.
Frequently asked questions
What substitution converts a family into its orthogonal trajectories?
Replace dy/dx by −dx/dy (the negative reciprocal) in the family's differential equation, then integrate the resulting equation.
What are the orthogonal trajectories of the circles x² + y² = c?
The straight lines y = kx through the origin: each circle meets each line radially, and radial directions are perpendicular to circular directions.
Why must the arbitrary constant be eliminated before substituting?
Because the trajectory condition applies to the whole family's slope field at each point; keeping c ties the answer to one member and produces a wrong, parameter-dependent family.
What are the trajectories of y = ce^x?
Differentiate to y' = y, invert to y' = −1/y, integrate to y² = −2x + C: leftward-opening parabolas.
How do you verify two families are orthogonal?
Pick any intersection point, compute both slopes there, and check their product is −1.