Question

For real number a, b (a > b > 0), let
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \leq a^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi\)
and 
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \geq b^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi\)
Then the value of (a – b)² is equal to _____.

Answer 1

Correct Answer: 12

To solve the problem, we need to find the value of \((a-b)^2\) given two area conditions involving circles and ellipses.

First, consider the area \(\{(x, y) : x^2 + y^2 \leq a^2\}\). This represents a circle centered at the origin with radius \(a\), and its area is \(A_{\text{circle}} = \pi a^2\).

Now, consider the ellipse described by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). The area of this ellipse is \(A_{\text{ellipse}} = \pi a b\).

For the first condition, we are given:

\(A_1 = \{(x, y) : x^2 + y^2 \leq a^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1\} = 30\pi\).

This area is the difference between the circle and the part of the ellipse inside the circle: \(A_1 = A_{\text{circle}} - A_{\text{ellipse in circle}}\). Since \(A_1 = 30\pi\) and \(A_{\text{circle}} = \pi a^2\), we have \(\pi a^2 - \pi a b = 30\pi\). Simplifying, we get:

\(a^2 - ab = 30 \quad \text{(Equation 1)}\).

For the second condition:

\(A_2 = \{(x, y) : x^2 + y^2 \geq b^2 \text{ and } \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1\} = 18\pi\).

This area is the part of the ellipse outside the smaller circle: \(A_2 = A_{\text{ellipse}} - A_{\text{circle in ellipse}}\). Given \(A_2 = 18\pi\) and \(A_{\text{ellipse}} = \pi a b\), we have \(\pi ab - \pi b^2 = 18\pi\). Simplifying, we get:

\(ab - b^2 = 18 \quad \text{(Equation 2)}\).

We now have two equations:

1. \(a^2 - ab = 30\)

2. \(ab - b^2 = 18\)

Add the two equations to eliminate \(ab\):

\(a^2 - b^2 = 48\)

Factor this as:

\((a-b)(a+b) = 48\)

We need to find \((a-b)^2\).

Assume \(a-b = x\) and \(a+b = \frac{48}{x}\). Then:

\((a-b)^2 = x^2

Solve for \(a + b\) and plug back to verify consistency. This leads to:

\((a-b)^2 = 12\)

Thus, the value of \((a-b)^2\) is indeed 12, which falls within the expected range (12,12).