# Orthogonal Trajectories > Orthogonal trajectories for JEE Mathematics: replace dy/dx by negative reciprocal, solve the new differential equation, worked family examples. - Canonical URL: https://prepelephant.com/topics/jee/mathematics/orthogonal-trajectories - Exam / course: JEE · Subject: Mathematics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Orthogonal Trajectories", PrepElephant, https://prepelephant.com/topics/jee/mathematics/orthogonal-trajectories ## 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.