1. Tardigrade
  2. Question
  3. Mathematics
  4. For real numbers a, b (a> b >0), let Area (x, y): x2+y2 ≤ a2. and .(x2/a2)+(y2/b2) ≥ 1 =30 π and Area (x, y): x2+y2 ≥ b2. and .(x2/a2)+(y2/b2) ≤ 1 =18 π Then the value of (a- b )2 is equal to .

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. For real numbers $a, b (a> b >0)$, let
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 MainJEE Main 2022Application of Integrals

Solution:

given $\pi a ^{2}-\pi ab =30 \pi$ and $\pi ab -\pi b ^{2}=18 \pi$
on subtracting, we get $(a-b)^{2}=a^{2}-2 a b+b^{2}=12$