Let y = p ( x ) be the parabola passing through the points ( - 1 , 0 ) , ( 0 , 1 ) and ( 1 , 0 ) . If the area of the region { ( x , y ) : ( x + 1 ) 2 + ( y - 1 ) 2 1 , y p ( x ) } is A , then 12 ( - 4 A ) is equal to ______ . [2023]

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Solution

Q.
Let $y = p (x)$ be the parabola passing through the points $(- 1, 0), (0, 1)$ and $(1, 0)$ . If the area of the region

${(x, y) : {(x + 1)}^{2} + {(y - 1)}^{2} \leq 1, y \leq p (x)}$ is $A$ , then $12 (π - 4 A)$ is equal to ______ . [2023]

Ans.
(16)

There can be infinite parabola through given points.

In question, it must be given that axis of parabola is parallel to $y$ -axis.

Equation of parabola passing through (-1, 0), (0, 1) and (1, 0) is $x^{2} = - (y - 1)$ ...(i)

$∴$ $Required area, A = \int_{- 1}^{0} [(1 - x^{2}) - (1 - \sqrt{1 - {(x + 1)}^{2}})] d x$

$= \int_{- 1}^{0} (- x^{2} + \sqrt{1 - {(x + 1)}^{2}}) d x$

$= {[\frac{- x^{3}}{3} + \frac{x + 1}{2} \sqrt{1 - {(x + 1)}^{2}} + \frac{1}{2} \sin^{- 1} (\frac{x + 1}{1})]}_{- 1}^{0}$

$= \frac{1}{2} (\frac{π}{2}) - \frac{1}{3} = \frac{π}{4} - \frac{1}{3}$

$∴$ $12 (π - 4 A) = 12 (π - 4 (\frac{π}{4} - \frac{1}{3})) = 12 (\frac{4}{3}) = 16$

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