---
title: "Tangent at (a,b) to (x/a)³+(y/b)³=2"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:tangent-at-a-b-cubic-curve"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/tangent-at-a-b-cubic-curve
site: BCT Engineering
---

# Tangent at (a,b) to (x/a)³+(y/b)³=2

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2069 Asar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/tangent-at-a-b-cubic-curve
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Find the tangent at (a, b) to the curve (x/a)³ + (y/b)³ = 2.

## Short answer

Implicit differentiation gives slope –b/a at (a,b), so the tangent is x/a + y/b = 2, i.e., bx + ay = 2ab.

## Answer

Curve: (x/a)³ + (y/b)³ = 2. The point (a, b) lies on it, since 1 + 1 = 2.

**Slope.** Differentiate implicitly:

3x²/a³ + 3y² (dy/dx)/b³ = 0, so dy/dx = −(b³ x²)/(a³ y²).

At (a, b): dy/dx = −(b³ a²)/(a³ b²) = −b/a.

**Tangent:** y − b = −(b/a)(x − a). Multiply by 1/b:

(y − b)/b = −(x − a)/a, so x/a + y/b = 2.

**Answer: the tangent at (a, b) is x/a + y/b = 2, i.e. bx + ay = 2ab.**

## Check yourself

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

1. The slope of the tangent is found by
   A. implicit differentiation and substituting (a, b)
   B. Substituting (a, b) before differentiating
   C. Finding dx/dy and taking the reciprocal at x = 0
   D. Using the chord joining the origin to (a, b)
   Answer: A. Differentiate implicitly to get dy/dx, then evaluate at the point. (Key point: Method)

2. The slope at (a, b) is
   A. b/a
   B. -b/a
   C. -a/b
   D. a/b
   Answer: B. dy/dx = -(b^3 x^2)/(a^3 y^2) = -(b^3 a^2)/(a^3 b^2) = -b/a. (Key point: Slope)

3. The tangent at (a, b) is
   A. x/a + y/b = 1
   B. x/a - y/b = 0
   C. x/a + y/b = 2
   D. ax + by = 2
   Answer: C. y - b = -(b/a)(x - a) gives x/a + y/b = 2. (Key point: Tangent equation)

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