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

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

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2072 Chaitra; 2069 Asar (asked 2 times)
- 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-a-minus-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the area enclosed by y²(a−x) = x³ and its asymptotes.

## Short answer

The asymptote is x=a. Area = 2∫₀^a x^(3/2)(a−x)^(−1/2) dx. Substitute x = a sin²θ to obtain 4a² ∫₀^{π/2} sin⁴θ dθ = 3πa²/4.

## Answer

Curve: y²(a − x) = x³ (the cissoid of Diocles). Then y² = x³/(a − x).

- The curve exists for 0 ≤ x < a (for x < 0 or x > a the right side is negative).
- It is symmetric about the x-axis, passes through the origin (a cusp), and as x → a, y → ±∞. So **x = a** is the asymptote.

The required area lies between the curve and the asymptote x = a, so it is twice the area above the x-axis:

```text
A = 2 ∫₀^a y dx = 2 ∫₀^a x^(3/2) (a − x)^(−1/2) dx
```
Put x = a sin²θ, dx = 2a sin θ cos θ dθ, a − x = a cos²θ. Limits: 0 to π/2.

```text
A = 2 ∫₀^(π/2) a^(3/2) sin³θ · a^(−1/2) (1/cosθ) · 2a sinθ cosθ dθ
  = 4a² ∫₀^(π/2) sin⁴θ dθ
  = 4a² · (3/4)(1/2)(π/2)
  = 4a² · 3π/16
```
**Area = 3πa²/4.**

## Check yourself

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

1. The asymptote of y^2(a-x) = x^3 is
   A. y = a
   B. x = 0
   C. x = a
   D. y = x
   Answer: C. As x → a, y → ±∞. (Key point: Asymptote)

2. The area is
   A. integral from 0 to a of x^3/(a-x) dx
   B. 2 integral from -a to a of y dx
   C. integral from 0 to ∞ of y dx
   D. 2 integral from 0 to a of x^(3/2)(a-x)^(-1/2) dx
   Answer: D. Twice the area above the x-axis between curve and asymptote. (Key point: Setup)

3. The substitution used is
   A. x = a sin^2 θ
   B. x = a tan θ
   C. x = a cos θ
   D. x = a θ
   Answer: A. Then a - x = a cos^2 θ. (Key point: Substitution)

4. The area is
   A. 3πa^2
   B. 3πa^2/4
   C. πa^2/4
   D. 3πa^2/8
   Answer: B. 4a^2 · 3π/16 = 3πa^2/4. (Key point: Result)

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