---
title: "Radius of curvature of x³+y³=3axy at origin"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:curvature-origin-folium"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-origin-folium
site: BCT Engineering
---

# Radius of curvature of x³+y³=3axy at origin

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2074 Asoj; 2073 Shrawan; 2071 Shrawan, 5 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/curvature-origin-folium
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Find the radius of curvature of the curve x³ + y³ = 3axy at origin.

## Short answer

The curve x³+y³=3axy has a node at the origin with tangents x=0, y=0. For the branch tangent to the x-axis, y ≈ x²/(3a) and radius of curvature ρ = lim x²/(2y) = 3a/2. Symmetry gives the same for the other branch.

## Answer

For a curve passing through the origin, the formula of Newton is used: if the x-axis is the tangent at the origin, then ρ = lim_{x→0} x²/(2y); if the y-axis is the tangent, ρ = lim y²/(2x).

The curve x³ + y³ = 3axy has the lowest-degree terms 3axy = 0 at the origin, so the tangents at the origin are x = 0 and y = 0 (a node).

**Tangent y = 0 (x-axis):** near the origin y is small, so y³ is negligible compared with the other terms. Then x³ ≈ 3axy, giving y ≈ x²/(3a). Check: let y = c₂x² + ..., then x³ − 3a c₂ x³ = 0 gives c₂ = 1/(3a).

ρ = lim x²/(2y) = x²/(2x²/(3a)) = 3a/2

(Equivalently ρ = (1+y′²)^(3/2)/y″ with y′ = 0, y″ = 2/(3a).)

**Tangent x = 0 (y-axis):** by symmetry in x and y, x ≈ y²/(3a), and ρ = lim y²/(2x) = 3a/2.

**Answer: the radius of curvature at the origin is 3a/2 (for both branches).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. If the x-axis is the tangent at the origin, the radius of curvature is:
   A. lim 2y/x^2
   B. lim y^2/(2x)
   C. lim x^2/(2y)
   D. lim x/y
   Answer: C. For x-axis tangent, rho = lim x^2/(2y); for y-axis tangent it is lim y^2/(2x). (Key point: Newton's formula)

2. Why are x = 0 and y = 0 the tangents at the origin of x^3 + y^3 = 3axy?
   A. The lowest-degree term 3axy equated to zero gives them
   B. The curve is symmetric
   C. y' = 0 at the origin
   D. x^3 + y^3 = 0 there
   Answer: A. Equating the lowest-degree terms to zero gives the tangents at the origin. (Key point: Tangents at origin)

3. Near the origin on the x-axis branch, what is y approximately?
   A. 3a x^2
   B. x^3/(3a)
   C. x^2/(3a)
   D. 3a/x
   Answer: C. With y^3 negligible, x^3 = 3axy gives y = x^2/(3a). (Key point: Approximation)

4. What is the radius of curvature at the origin?
   A. 2a/3
   B. a/3
   C. 3a/2
   D. 3a
   Answer: C. rho = x^2/(2 * x^2/(3a)) = 3a/2, and the same for the other branch by symmetry. (Key point: Result)

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