---
title: "Area between y²(2a−x)=x³ and its asymptote"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:area-asymptote-curve-y2-2a-minus-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/area-asymptote-curve-y2-2a-minus-x
site: BCT Engineering
---

# Area between y²(2a−x)=x³ and its asymptote

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2068 Chaitra
- Video: [Find the area between \( y^{2}=\frac{x^{3}}{a-x} \) and its asymptote.](https://www.youtube.com/watch?v=696_ZzYcwfE) — PW Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/area-asymptote-curve-y2-2a-minus-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the area of the curve y²(2a−x) = x³ and its asymptotes.

## Short answer

Asymptote is x=2a. Area = 2∫₀²ᵃ x^(3/2)/√(2a−x) dx. Substituting x=2a sin²θ reduces to 16a²∫₀^{π/2} sin⁴θ dθ = 3πa².

## Answer

Curve: y²(2a − x) = x³ (cissoid), so y² = x³/(2a − x). It is symmetrical about the x-axis, exists for 0 ≤ x < 2a, passes through the origin and has the asymptote **x = 2a** (y → ±∞ as x → 2a).

The area between the curve and the asymptote is twice the area above the x-axis:

A = 2 ∫₀^(2a) x^(3/2) (2a − x)^(−1/2) dx

Put x = 2a sin²θ, dx = 4a sin θ cos θ dθ, 2a − x = 2a cos²θ. Limits 0 to π/2.

```text
A = 2 ∫₀^(π/2) (2a)^(3/2) sin³θ · (2a)^(−1/2) (1/cosθ) · 4a sinθ cosθ dθ
  = 2 · 2a · 4a ∫₀^(π/2) sin⁴θ dθ
  = 16a² · 3π/16
```
(using ∫₀^(π/2) sin⁴θ dθ = (3/4)(1/2)(π/2) = 3π/16)

**Area = 3πa².**

## Check yourself

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

1. The asymptote is
   A. x = a
   B. x = 2a
   C. y = 2a
   D. x = 0
   Answer: B. y → ±∞ as x → 2a. (Key point: Asymptote)

2. Use
   A. x = 2a tan θ
   B. x = a sin^2 θ
   C. x = 2a sin^2 θ
   D. x = 2a cos θ
   Answer: C. Then 2a - x = 2a cos^2 θ. (Key point: Substitution)

3. The area reduces to
   A. 8a^2 integral sin^2 θ dθ
   B. 4a^2 integral sin^4 θ dθ
   C. 16a^2 integral sin^3 θ dθ
   D. 16a^2 integral sin^4 θ dθ over 0 to π/2
   Answer: D. 2 · 2a · 4a = 16a^2. (Key point: Reduction)

4. The area is
   A. 3πa^2
   B. 3πa^2/4
   C. 3πa^2/2
   D. πa^2
   Answer: A. 16a^2 · 3π/16. (Key point: Result)

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