For real numbers a, b (a b 0), let Area { (x, y) : x² + y² ≤ a² and x²/a² + y²/b² ≥ 1 } = 30π an

Q: Wait, do I need to solve for $a$ and $b$ individually to find $(a - b)^2$?

That is a classic trap! Many students waste precious minutes trying to isolate $a$ and $b$ by solving the system of equations. While you *could* do that, notice that the question asks for $(a - b)^2$. In competitive exams, always look at the target expression first. You will find that the algebraic structure allows you to reach the answer without ever knowing the individual values of $a$ and $b$.

Q: Why is the area of the ellipse $\pi ab$? I thought it was related to integration.

You are absolutely right that it comes from integration! The area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is derived by integrating $y = b\sqrt{1 - \frac{x^2}{a^2}}$ from $-a$ to $a$. However, for JEE Advanced, you should treat $\pi ab$ as a standard formula in your toolkit, just like $\pi r^2$ for a circle. Memorizing this saves you time and reduces the chance of calculation errors.

Q: Why does the inequality $\frac{x^2}{a^2} + \frac{y^2}{b^2} \ge 1$ represent the region 'outside' the ellipse?

Think about the boundary. The equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is the boundary of the ellipse. If you pick a point far away from the origin, say $(10a, 10b)$, the value of the expression $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ will be much larger than 1. Conversely, points inside the ellipse yield values less than 1. Thus, $\ge 1$ is indeed the exterior.

Question

For real numbers $a, b (a > b > 0)$, let $Area \left\{ (x, y) : x^2 + y^2 \le a^2 	ext{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \ge 1 ight\} = 30\pi$ and $Area \left\{ (x, y) : x^2 + y^2 \ge b^2 	ext{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 ight\} = 18\pi$. Then the value of $(a - b)^2$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, coordinate-plane workshop. Before you lie three distinct geometric shapes, all centered perfectly at the origin. We have an outer circle, an ellipse, and an inner circle.

This is not just a math problem; it is a study of nested worlds. We are given the condition $a > b > 0$. This inequality is our compass, indicating that the outer circle with radius $a$ is the largest boundary, the ellipse with semi-axes $a$ and $b$ sits inside it, and the inner circle with radius $b$ is tucked inside the ellipse.

### Translating Geometry into Language

Let us translate these shapes into the language of mathematics:

*   Outer Circle: $x^2 + y^2 = a^2$
*   Ellipse: 
    $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
*   Inner Circle: $x^2 + y^2 = b^2$

The first region is the area inside the outer circle but outside the ellipse. The area of this region is the area of the outer circle minus the area of the ellipse:
$$ \pi a^2 - \pi ab = 30\pi $$

Dividing by $\pi$, we obtain our first equation:
$$ a^2 - ab = 30 $$

### The Second Layer

Now, let us turn our attention to the second region: the area inside the ellipse but outside the inner circle. This is calculated as the area of the ellipse minus the area of the inner circle:
$$ \pi ab - \pi b^2 = 18\pi $$

Dividing by $\pi$ once more, we arrive at our second equation:
$$ ab - b^2 = 18 $$

### The Algebraic Finale

We now have two powerful tools in our arsenal:
1) $a^2 - ab = 30$
2) $ab - b^2 = 18$

The question asks us to find the value of $(a - b)^2$. Recall the algebraic expansion:
$$ (a - b)^2 = a^2 - 2ab + b^2 $$

If we subtract the second equation from the first, we get:
$$ (a^2 - ab) - (ab - b^2) = 30 - 18 $$

Expanding the left side yields:
$$ a^2 - ab - ab + b^2 = a^2 - 2ab + b^2 $$

This is exactly the expansion of $(a - b)^2$. On the right side, $30 - 18$ gives us $12$.

The final result is **12**.

Enter Numerical Value:

Visualized Solution

Setting the Stage

Outer Circle and Ellipse

Inner Circle

Recalling Area Formulas

Analyzing Region 1

Formulating Equation 1

Analyzing Region 2

Formulating Equation 2

The Target Expression

Subtracting the Equations

Simplifying to the Final Answer

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Translating Geometry into Language

The Second Layer

The Algebraic Finale

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