Cutting a Parabolic Region in Half

Geometry Level pending

A parabolic region is bounded by the curve y = a x 2 y = a x^2 , and y = b y = b . Where should we draw a horizontal line that will cut this region by half ? If the horizontal line is y = c y = c , find c c .

2 3 2 b 2^{-\frac{3}{2}} b 2 1 b 2^{-1} b 2 2 3 b 2^{- \frac{2}{3}} b 2 1 2 b 2^{-\frac{1}{2} } b

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2 solutions

David Vreken
May 27, 2020

The line and curve y = b y = b and y = a x 2 y = ax^2 intersect at ( ± b a , b ) (\pm \sqrt{\frac{b}{a}}, b) , and since the area of a parabola is 2 3 \frac{2}{3} the area of its inscribed rectangle, the area of the large parabola is A L = 2 3 2 b a b = 4 b b 3 a A_L = \frac{2}{3} \cdot 2\sqrt{\frac{b}{a}} \cdot b = \frac{4b\sqrt{b}}{3\sqrt{a}} .

Similarly, the area of the small parabola bounded by y = c y = c and y = a x 2 y = ax^2 is A S = 4 c c 3 a A_S = \frac{4c\sqrt{c}}{3\sqrt{a}} .

To cut the region in half, 2 A S = A L 2A_S = A_L , or 2 4 c c 3 a = 4 b b 3 a 2 \frac{4c\sqrt{c}}{3\sqrt{a}} = \frac{4b\sqrt{b}}{3\sqrt{a}} , and this solves to c = 2 2 3 b c = \boxed{2^{-\frac{2}{3}}b} .

From the given conditions of the problem we get 0 b y d y = 2 0 c y d y b 3 2 = 2 c 3 2 c = 2 2 3 b \displaystyle \int_0^b \sqrt y dy=2\displaystyle \int_0^c \sqrt y dy\implies b^{\frac{3}{2}}=2c^{\frac{3}{2}}\implies c=\boxed {2^{-\frac{2}{3}}b} .

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