trace the curve ay^2=x^2(a-x) and find the area of its loop

Let's dive into tracing the curve given by the equation ay² = x²(a - x) and then determine the area enclosed by its loop. This is a classic problem involving symmetry and integration, often asked in advanced calculus or coordinate geometry sections of IIT JEE-level exams.

Step-by-Step Curve Analysis

1. Identify the Type and Symmetry

Given equation: ay² = x²(a - x)

• This is a Cartesian equation relating x and y.
• Since y appears only as y², the curve is symmetric about the x-axis.

2. Domain of the Curve (Where y² ≥ 0)

Since y² must be non-negative:

⇒ ay² = x²(a - x) ≥ 0

We analyze when the right-hand side is non-negative:

• x² is always non-negative.
• So the sign of the RHS depends on (a - x).

Thus, x ∈ [0, a] for the curve to exist (because beyond x = a, RHS becomes negative).

3. Points of Intersection with Axes

• At x = 0: ay² = 0 → y = 0 ⇒ Point (0, 0)
• At x = a: ay² = a²(0) = 0 → y = 0 ⇒ Point (a, 0)

So, the curve starts and ends on the x-axis at x = 0 and x = a.

4. Shape and Behavior

This curve resembles a loop that lies entirely between x = 0 and x = a, and is symmetric about the x-axis.

Find the Area Enclosed by the Loop

Set up the Integral

Given: ay² = x²(a - x) ⇒ \( y = \sqrt{ \frac{x^2(a - x)}{a} } \)

The total area is given by:

So,

Take \( \frac{1}{\sqrt{a}} \) outside the integral:

Substitute to Solve the Integral

Let \( u = a - x \) ⇒ \( du = -dx \), and when x = 0, u = a; when x = a, u = 0

Then:

So the integral becomes:

Now expand the integrand:

Integrate:

Now simplify:

Compute the bracket:

Now multiply:

Final Answer

The area enclosed by the loop of the curve ay² = x²(a - x) is: