QJEEM22Ma1cae: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1cae Solution: For real numbers a, b (a > b > 0), let Area { ( x , y ) : x 2 + …

Question

Verified Solution • 2025–26 Session moderate

For real numbers a, b (a > b > 0), let

Area ${(x, y) : x^{2} + y^{2} \leq a^{2} a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \geq 1} = 30 π$

and

Area ${(x, y) : x^{2} + y^{2} \leq b^{2} a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1} = 18 π$

Then, the value of (a $-$ b)² is equal to ___________.

Accepted Answer

x^{2}+y^{2} \leq a^{2}

is interior of circle

and

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

is exterior of ellipse

JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Area Under The Curves Question 79 English Explanation

\therefore

Area

=\pi a^{2}-\pi a b=30 \pi

Similarly

x^{2}+y^{2} \geq b^{2}

and

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

gives

\pi a b-\pi b^{2}=18 \pi

By (1) and (2),

\frac{a}{b}=\frac{5}{3} \Rightarrow a=\frac{5 b}{3}

\Rightarrow \pi \cdot \frac{25 b^{2}}{9}-\pi \cdot \frac{5 b^{2}}{3}=30 \pi

\Rightarrow\left(\frac{25}{9}-\frac{5}{3}\right) b^{2}=30

\Rightarrow \frac{10}{9} b^{2}=30 \Rightarrow b^{2}=27

and

a^{2}=\frac{25}{9} \cdot 27=75

(a-b)^{2}=(5 \sqrt{3}-3 \sqrt{3})^{2}=3 \cdot 4=12

Step-by-Step Verified Solution