For real numbers \(a, \mathrm{~b}(a>\mathrm{b}>0)\), let \(\operatorname{Area}\left\{(x, y): x^{2}+y^{2} \leq a^{2}\right.\) and \(\left.\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \geq 1\right\}=30 \pi\) and Area \(\left\{(x, y): x^{2}+y^{2} \geq b^{2}\right.\) and \(\left.\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1\right\}=18 \pi\) Then the value of \((a-b)^{2}\) is equal to ______.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Area Under Curves

For real numbers

a, b (a > b > 0)

, let

Area {(x, y) : x^{2} + y^{2} \leq a^{2}

and

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} \geq 1} = 30 π

and
Area

{(x, y) : x^{2} + y^{2} \geq b^{2}

and

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} \leq 1} = 18 π

Then the value of

(a - b)^{2}

is equal to ______.

Answer: 12

The first condition describes the region inside the circle

x^{2} + y^{2} = a^{2}

but outside the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

. This area is

π a^{2} - π ab = 30 π

.

The second condition describes the region inside the ellipse but outside the circle

x^{2} + y^{2} = b^{2}

. This area is

π ab - π b^{2} = 18 π

.

Subtracting the second equation from the first:

π a^{2} - π ab - (π ab - π b^{2}) = 30 π - 18 π

π a^{2} - 2 π ab + π b^{2} = 12 π

π (a^{2} - 2 ab + b^{2}) = 12 π

(a - b)^{2} = 12

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer