For real numbers a, b with a b 0, the area of the region {(x,y): x² + y² = 1} equals 30*pi, and the area of the region {(x,y): x² + y² = b² and x²/a² + y²/b²

Question

For real numbers a, b with a > b > 0, the area of the region {(x,y): x² + y² <= a² and x²/a² + y²/b² >= 1} equals 30*pi, and the area of the region {(x,y): x² + y² >= b² and x²/a² + y²/b² <= 1} equals 18*pi. Find (a - b)².

Accepted Answer

4. Region 1: inside circle (radius a) AND outside ellipse. Area = pi*a² - pi*a*b = pi*a(a-b) = 30*pi => a(a-b) = 30. Region 2: inside ellipse AND outside circle (radius b). Area = pi*a*b - pi*b² = pi*b(a-b) = 18*pi => b(a-b) = 18. Dividing: a/b = 30/18 = 5/3. Let a = 5k, b = 3k. Then b(a-b) = 3k*2k = 6k² = 18 => k² = 3 => k = sqrt(3). a = 5*sqrt(3), b = 3*sqrt(3). (a-b)² = (2*sqrt(3))² = 4*3 = 12. Hmm, this gives 12 which is not in the options. Let me try again: a(a-b)=30, b(a-b)=18 => (a-b)(a+b)... no, just divide: a/b = 5/3. a=5t, b=3t. b(a-b)=3t*2t=6t²=18 => t²=3. (a-b)²=(2t)²=4t²=12. Not in options. Alternative: maybe area of ellipse sector needs reconsideration. If options include 12, a

For real numbers a, b with a > b > 0, the area of the region {(x,y): x² + y² <= a² and x²/a² + y²/b² >= 1} equals 30pi, and the area of the region {(x,y): x² + y² >= b² and x²/a² + y²/b² <= 1} equals 18pi. Find (a - b)².

Solution

Related JEE Main Maths questions