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  • Let y = p(x) be the parabola passing through the points (–1, 0), (0, 1) and (1, 0). If the area of the region <img alt="\left \{ (x,y):(x+1)^{2}+(y-1)^{2}\leq 1,y\leq p(x) \right \}" src

Let y = p(x) be the parabola passing through the points (–1, 0), (0, 1) and (1, 0). If the area of the region \left \{ (x,y):(x+1)^{2}+(y-1)^{2}\leq 1,y\leq p(x) \right \}    is A, then  12(\pi-4A)  is equal to _____:

Option: 1

16


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

There can be infinitely many parabolas through given points.
Let parabola  x^{2}=-4a(y-1)

                                            

Passes through (1, 0)

\therefore b+-4a(-1)\Rightarrow a+\frac{1}{4}

\therefore x^{2}=-(y-1)

Now area covered by parabola      =\int_{-1}^0\left(1-x^2\right) d x

\begin{aligned} & =\left(x-\frac{x^3}{3}\right)_1^b=(0-0)-\left\{-1+\frac{1}{3}\right\} \\ & =\frac{2}{3} \end{aligned}

Required Area = Area of sector – {Area of square – Area covered by Parabola}

\begin{aligned} & =\frac{\pi}{4}-\left\{1-\frac{2}{3}\right\} \\ & =\frac{\pi}{4}-\frac{1}{3} \\ & \therefore 12(\pi-4 \mathrm{~A})=12\left[\pi-4\left(\frac{\pi}{4}-\frac{1}{3}\right)\right] \\ & =12\left[\pi-\pi+\frac{4}{3}\right] \\ \end{aligned}

=16

 

Posted by

shivangi.shekhar

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