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For real numbers \mathrm{a, b(a>b>0)} let

\mathrm{\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

\mathrm{\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 \mathrm{(a-b)^{2}} is equal to _______.

Option: 1

12


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{x^{2}+y^{2} \leq a} represents region inside circle  \mathrm{x^{2}+y^{2}=a^{2}} and

\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \geqslant 1} represents region outside the ellipse  \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}

First area  \mathrm{=30\pi=} area of circle - area of ellipse

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

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

\mathrm{\Rightarrow a^{2}-a b=30 }                    .........(i)

Similarly second area  \mathrm{=18 \pi=\pi a b-\pi b^{2}}

\mathrm{\Rightarrow \quad a b-b^{2}=18}                     ..........(ii)

Subtracting (i)  -  (ii)

\mathrm{\Rightarrow a^{2}-a b-a b+b^{2}=30-18} \\

\mathrm{\Rightarrow(a-b)^{2}=12 }

Hence answer is \mathrm{12 }

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mansi

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